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2025-03-21T14:48:29.915502
| 2020-02-20T00:14:20 |
353119
|
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|
Stack Exchange
|
How general are Gröbner degenerations?
While working with flat degenerations of flag varieties and Schubert varieties I've noticed that among the numerous known constructions there doesn't seem to be a single one that doesn't turn out to be a Gröbner degeneration. I've started asking myself if there's a concrete mathematical reason for this. In general terms my question is: is there some reasonably large class of situations in which all flat degenerations are Gröbner degenerations? To pose a precise question I will use a very simple and perhaps naive setting (mostly because I'm not really an algebraic geometer).
Consider an ideal $I$ in the polynomial ring $R=\mathbb C[x_1,\dots,x_n,t]$. I will assume that $I$ is homogeneous in the $x_i$ (but not $t$). To keep things nice and geometric I will also require $I$ to be a radical ideal. Denote $R_0=\mathbb C[x_1,\dots,x_n]$, then for every complex $c$ we have an ideal $I(c)\subset R_0$ which is obtained from $I$ by setting $t=c$. Suppose that the algebras $R_0/I(c)$ are pairwise isomorphic for all $c\neq 0$. The composition map in $\mathbb C[t]\hookrightarrow R\twoheadrightarrow R/I$ turns $R/I$ into a $\mathbb C[t]$-module, suppose this module is flat. (What I'm going for is a flat family of projective subvarieties in $\mathbb P^{n-1}$ over $\mathbb A^1$ with a constant fiber outside of $0$.)
Let me call the above construction a "flat family". In particular, let $I_1\subset R_0$ be a homogeneous ideal and $I_0\subset R_0$ is a (radical) initial ideal of $I_1$ with respect to some monomial order or grading on $R_0$. There's a standard way of constructing a flat family with $I(1)=I_1$, $R/I(c)\simeq R/I_1$ for $c\neq 0$ and $I(0)=I_0$. I will call this a "Gröbner family".
Is there an example of a flat family that is not (at least fiberwise) isomorphic to a Gröbner family?
Is there some natural set of conditions under which a flat family is necessarily (fiberwise?) isomorphic to a Gröbner family?
Update. I have realized that a simple (and not very interesting) way of obtaining flat families that are not Gröbner per se is to take a Gröbner degeneration and apply, say, a linear change of variables. This led me to give the questions an "up to isomorphism" flavor. It's been a year but I'm still very curious about this and still have no idea.
You would probably want it up to a "local" isomorphism, since any flat family with more than one special fiber would be a trivial counter example.
@auniket Yes, that is why I give a very restrictive definition of a flat family which, in particular, requires exactly one special fiber. Although I must say I don't know any examples with multiple special fibers either (where for at least one special fiber we don't have a Gröbner family such that...).
|
2025-03-21T14:48:29.915689
| 2020-02-20T00:29:22 |
353120
|
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|
Stack Exchange
|
Volume of solution sets for polynomials in $\mathbb{C}[x]$
Denote $\pmb{a}=(a_1,\dots,a_d)\in\mathbb{R}^d$ and consider the set
$$\mathcal{E}_d=\{\pmb{a}\in\mathbb{R}^d: \text{each root $\xi$ of $x^d+a_dx^{d-1}+\cdots+a_2x+a_1=0$ lies in $\vert\xi\vert<1$}\}.$$
In the reference shown below, Fam proved that the $d$-dimensional Lebesgue measure satisfies
$$\lambda_d(\mathcal{E}_d)=2^d\prod_{k=1}^{\lfloor\frac{d}2\rfloor}\left(1+\frac1{2k}\right)^{2k-d}.$$
I'd like to propose a complex version here. Denote $\pmb{c}=(c_1,\dots,c_d)\in\mathbb{C}^d$ and consider the set
$$\mathcal{S}_d=\{\pmb{c}\in\mathbb{C}^d: \text{each root $\xi$ of $x^d+c_dx^{d-1}+\cdots+c_2x+c_1=0$ lies in $\vert\xi\vert<1$}\}.$$
Now, let's ask:
QUESTION. What is the $2d$-dimensional Lebesgue measure
$$\lambda_{2d}(\mathcal{S}_d)?$$
For contrast, $\lambda_1(\mathcal{E}_1)=2$ while $\lambda_2(\mathcal{S}_1)=\pi$.
Reference.
A. T. Fam,The volume of the coefficient space stability domain of monic polynomials, Proc. IEEE Int. Symp.Circuits and Systems, 2 (1989), pp. 1780–1783.
$\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}$The answer is $\tfrac{\pi^n}{n!}$. It is certainly a surprise to have the answer come out so simple!
Let $\phi : \CC^n \to \CC^n$ be the map which takes $(z_1, z_2, \ldots, z_n)$ to the elementary symmetric functions $(e_1, e_2, \ldots, e_n)$ where $e_k = \sum_{1 \leq i_1 < i_2 < \cdots < i_k \leq n} z_{i_1} z_{i_2} \cdots z_{i_n}$. Let $D$ be the unit disc in $\CC$. So you want to compute the volume of $\phi(D^n)$, also known as $\int_{\phi(D^n)} \mathrm{Vol}$.
The map $D^n \to \phi(D^n)$ is $n!$ to $1$ and, since $\phi$ is complex analytic, $\phi$ is orientation preserving. So
$$\int_{\phi(D^n)} \mathrm{Vol} = \frac{1}{n!} \int_{D^n} \phi^{\ast}(\mathrm{Vol}) = \frac{1}{n!} \int_{D^n} \det J^{\RR}_{\phi}$$
where $J^{\RR}_{\phi}$ is the Jacobian of $\phi$ considered as a smooth map $\RR^{2n} \to \RR^{2n}$. I will write $J^{\CC}_{\phi}$ when I instead want the $n \times n$ matrix of complex numbers coming from thinking of $\phi$ as a complex analytic map $\CC^n \to \CC^n$.
The relation between these two notions is this: $\det J^{\RR}_{\phi} = |\det J^{\CC}_{\phi}|^2$. (This is just linear algebra -- if $L/K$ is a degree $d$ field extension, $f: L^n \to L^n$ is a linear map and $g: K^{dn} \to K^{dn}$ is the linear map gotten by identifying $L$ with $K^d$, then $\det g = N_{L/K}(\det f)$.) We have the well known identity
$$\det J^{\CC}_{\phi}(z_1, \ldots, z_n) = \prod_{i<j} (z_i - z_j) = \sum_{w \in S_n} (-1)^w z_1^{w(1)-1} z_2^{w(2)-1} \cdots z_n^{w(n)-1}.$$
Here is the first reference I found.
So we need to compute
$$\frac{1}{n!} \int_{D^n} \sum_{u \in S_n} (-1)^u z_1^{u(1)-1} z_2^{u(2)-1} \cdots z_n^{u(n)-1} \overline{\sum_{v \in S_n} (-1)^v z_1^{v(1)-1} z_2^{v(2)-1} \cdots z_n^{v(n)-1}}.$$
We can distribute the product to get a sum of $(n!)^2$ terms of the form $\int_{D^n} \prod z_j^{a_j} \overline{z_j}^{b_j}$. If $a_j \neq b_j$, the integral on $z_j$ is zero, so we reduce to
$$\frac{1}{n!} \sum_{u \in S_n} \int_{D^n} (z_1 \overline{z_1})^{u(1)-1} (z_2 \overline{z_2})^{u(2)-1} \cdots (z_n \overline{z_n})^{u(n)-1}.$$
All $n!$ terms have the same integral, so we are reduced to the one integral
$$\int_{D^n} \prod (z_j \overline{z_j})^{j-1} = \prod_j \int_D (z \overline{z})^{j-1}.$$
Switching to polar coordinates,
$$\int_D (z \overline{z})^{j-1} = 2 \pi \int_{r=0}^1 r^{2j-1} dr = \frac{\pi}{ j}.$$
So the final answer is $\prod_{j=1}^n \tfrac{\pi}{j} = \tfrac{\pi^n}{n!}$.
This is cool. Two things came to mind while reading your response: (1) somewhere it seems you had $det(A)=\pmb{Pfaff}(A)^2$; (2) the integral you wrote $\int \det$ when restricted to the hyperspheres may be translated as the "degree of a map" - I wonder if this has some relation here. Note: $\pmb{Pfaff}$ stands for Pfaffian of a matrix.
(1) I don't think I have any Pfaffians. The key fact is that, if $A : \mathbb{C}^n \to \mathbb{C}^n$ is a $\mathbb{C}$-linear map and $A^{\mathbb{R}}$ is the same map considered as an $\mathbb{R}$-linear map $\mathbb{R}^{2n} \to \mathbb{R}^{2n}$ then $\det A^{\RR} = |\det A|^2$. A determinant equaling a square is reminiscient of a Pfaffian, but I don't see one here. (2) Well, I use that $\phi$ has degree $n!$. I'm not sure what else to say.
Cool. $\pi^n/n!$ is also the volume of $D^n$ in $\mathbb{C}^n / S_n$ just computed naively.
@Nate I noticed that but couldn't think what to say about it!
|
2025-03-21T14:48:29.915988
| 2020-02-21T20:30:39 |
353272
|
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|
Stack Exchange
|
Definition of random measures
Introducing the notion of a random measure, textbooks usually start with a locally compact second countable Hausdorff space. Where does this requirement come from?
I would like to have a motivation for this requirement. That is, I would like to define a random measure on a general measurable space $(X,\mathcal{B})$ simply as a kernel from a probability space to $(X,\mathcal{B})$. Additional structure of $(X,\mathcal{B})$ should then be motivated by e.g. counterintuitive examples. For example, in the book of Last and Penrose (2017) (link to pdf),
Exercise 2.5 yields that point measures usually do not have the representation with Dirac measures.
Are there other examples, (intuitive) motivations and reasons to use a locally compact (!) second countable (!) Hausdorff space?
Once a locally compact second countable Hausdorff space $(X,\mathcal{T})$ is given, the theory of random measures just considers measures $\mu$ that are locally finite, that is, $\mu(K) < \infty$ for all relatively compact sets $K \subset X.$ Therefore, I wonder the following:
Let $\mathcal{B} = \sigma(\mathcal{T})$ be the Borel-$\sigma$-algebra on $X,$ let $M_X$ be the set of all measures on $(X,\mathcal{B})$ and let $\mathcal{M}_X$ be the $\sigma$-algebra generated by the evaluation maps $\mu \mapsto \mu(B)$ for all $B \in \mathcal{B}.$
Is the set of all locally finite measures $\mathcal{M}_X$-measurable?
If you want a $\sigma$-algebra for the set of probability measures on a measurable space, you are looking for this: https://ncatlab.org/nlab/show/Giry+monad
Thanks for the link! Actually I already have a $\sigma$-algebra, that is, I implicitly consider the $\sigma$-algebra on the set of all measures on $(X,\mathcal{B})$ that is generated by the evaluation maps $\mu \mapsto \mu(B)$ for all $B\in\mathcal{B}$
I think it may depend in part on what one wants to do with the random measure. Do you have some examples of textbooks where you've seen this, so we could see what the context is?
I think sometimes people use more general random measures. For example, Poisson point process of excursions (of a Markov process) is a random measure with values in the Skorohod space, a Polish space which is not locally compact. That said, weak convergence of measures is much simpler on locally compact Polish spaces. And I think it gets very difficult to work with on non-separable metric spaces.
@ Nate Eldredge: e.g. Kallenberg's books (Random Measures (1974), https://www4.stat.ncsu.edu/~boos/library/mimeo.archive/ISMS_1974_963.pdf, Foundations of Modern Probability Theory (2002), Random Measures and Applications (2017) ) and also tDaley and Vere-Jones "Introduction to the Theory of Point Processes" Vol. II other textbooks just use $\mathbb{R}^k$ or certain subsets
I think it might be good to add the descriptive set theory tag to the list of tags.
Let $D$ be the set of Dirac measures. In fact, $D$ would constitut a reasonable event, regarding probabilistic considerations (since we want to talk about random measures). However, in general it is not an event.
Consider, e.g., $X = [0,1]$ with the cocountable $\sigma$-algebra $\mathcal{B} = \sigma (\{\{x\} : x \in \mathcal{X})$ and note that the $\sigma$-algebra $\mathcal{M}_X$ is countably determined. Then, there is no countable set system $\mathcal{E} \subset \mathcal{B}$ such that $\text{pr}_{\mathcal{E}}^{-1}\big(\text{pr}_{\mathcal{E}}(D)\big) = D$, where $\text{pr}_{\mathcal{E}} : M_X \to [0,\infty]^{\mathcal{E}}$ is the projection of the set functions on $\mathcal{E}.$ This is due to the fact that we can define a measure $\mu$ by $\mu(B) = 1,$ if $B^c$ is countable and else $\mu(B) = 0,$ where $B \in \mathcal{B}.$ Note, $\mu$ is in $\text{pr}_{\mathcal{E}}^{-1}\big(\text{pr}_{\mathcal{E}}(D)\big)$ since $\mu(B) \in \{0,1\}$ for all $B,$ but $\mu$ is not in $D$ since there is no $x\in X$ such that $\mu = \delta_x.$ Hence, $D$ is not $\mathcal{M}_X$-measurable.
Let $(X,\mathcal{T})$ be a locally compact second countable Hausdorff space, then the set of all locally finite measures on $(X,\mathcal{B})$ is $\mathcal{M}_X$-measurable.
To prove this, we first note that there is a countable basis $\mathcal{U}$ of $\mathcal{T}$ and a sequence $(G_k)_{k\in\mathbb{N}}\in\mathcal{T}^\mathbb{N}$ of relatively compact sets such that $G_k \uparrow X$ for $k \to \infty.$ Then, $\mathcal{E} :=\big\{ \bigcap \mathcal{O} : \mathcal{O} \subset \mathcal{U} \cup G(\mathbb{N}),\ |\mathcal{O}| < \infty \big\}$
is a countable and intersection stable generator of $\mathcal{B}.$ According to the measure uniqueness theorem, a locally finite measure $\mu$ is uniquely determined by $\mu|_{\mathcal{E}}$. Consequently, the set of all locally finite measures is countably determined, that is, $\mathcal{M}_X$-measurable.
Nice proof :) although perhaps a bit more should be said about why $\mu$ is in $\mathrm{pr}\mathcal{E}^{-1}(\mathrm{pr}\mathcal{E}(D))$: If we let $E_1$ be the intersection of all the sets in $\mathcal{E}$ with countable complement, and let $E_2$ be the union of all the countable sets in $\mathcal{E}$, then clearly $E_1\setminus E_2$ is non-empty, and for any $x\in E_1\setminus E_2$, $\mu$ agrees with $\delta_x$ on $\mathcal{E}$.
You are right, one has to be careful at this point. Thanks for completing the proof :)
By way of introduction:
As expressed in some of the comments, I find the "locally compact" assumption possibly a bit too strong.
A weaker assumption than having a locally compact second-countable Hausdorff space would be that the space is a Lusin space, i.e. a separable metrisable topological space that satisfies the following beautifully equivalent statements:
$X$ can be topologically identified with a Borel subset of a completely metrisable topological space;
for any metrisation $d$ of the topology of $X$, $X$ is a Borel subset of the $d$-completion of $X$;
if $X$ is uncountable then $(X,\mathcal{B}(X))$ is measurably isomorphic to $([0,1],\mathcal{B}([0,1]))$.
[I emphasise that this equivalence assumes that $X$ is a separable metrisable space.]
If $X$ is a Lusin space then we have some very nice properties in regards to random measures. But I will also explain some of the important properties more generally of second countable spaces, and of separable metrisable spaces.
1. Countably generated Borel space
The first thing to say is that everything is pretty much hopeless if you're working on a space $X$ for which the Borel $\sigma$-algebra $\mathcal{B}(X)$ is not countably generated. Much of probability theory is about almost sure statements, and it's often really important to be able to move logically from
"for each $A \in \mathcal{B}(X)$, some assertion is almost surely true"
to
"it's almost surely true that for every $A \in \mathcal{B}(X)$ the assertion holds".
The ability to do this often relies on $\mathcal{B}(X)$ being countably generated. This is fundamental to why second countability (or equivalently, for metrisable spaces, separability) is assumed - it guarantees that the Borel $\sigma$-algebra is countably generated.
2. Measurable structure on the space of probability measures
It would be really good if, on the space $M_X$ of Borel probability measures on $X$, the evaluation $\sigma$-algebra $\sigma(\mu \mapsto \mu(A) : A \in \mathcal{B}(X))$ is a "nice" $\sigma$-algebra to work with. We have the following:
Theorem 1. Let $X$ be a Lusin space (resp. any separable metrisable space). Then $M_X$ equipped with the topology of weak convergence is also a Lusin space (resp. a separable metrisable space), and the Borel $\sigma$-algebra of the topology of weak convergence is precisely the evaluation $\sigma$-algebra.
An immediate corollary is that if $X$ is a separable metrisable space then the evaluation $\sigma$-algebra is countably generated.
[Weak convergence of probability measures on a separable metric space has several equivalent definitions, one being that for every bounded continuous $g \colon X \to \mathbb{R}$, $\int g \, d\mu_n \to \int g \, d\mu$. The topology of weak convergence is a very nice and physically natural topology. One of its useful properties is that weak convergence can be determined using only countably many bounded continuous functions $g \colon X \to \mathbb{R}$.]
Sorry I don't have a good reference off hand for the above theorem, but I imagine it should be easy to find the result (maybe not stated as one single theorem) in textbooks or lecture notes on descriptive set theory and/or measure theory. The fact that the topology of weak convergence for a Lusin space is Lusin might not be stated explicitly, but what will probably be stated is that the topology of weak convergence for a compact metrisable space is compact; and since $[0,1]$ is obviously compact, combining this statement with all the other parts of the statement of Theorem 1 will then yield that the topology of weak convergence for a Lusin space is Lusin.
3. Disintegration of measures, and conditional expectation of random measures
For me, one of the marvels of Lusin spaces is the following disintegration theorem. Let $M_X$ be the set of Borel probability measures on $X$.
Theorem 2. Let $(\Omega,\mathcal{F},\mathbb{P})$ be an arbitrary probability space, and let $X$ be a Lusin space. Let $M_{\Omega,X}$ be the set of functions $\dot{\mu}\colon \Omega \to M_X$ such that $\omega \mapsto \dot{\mu}(\omega)(A)$ is an $\mathcal{F}$-measurable map for all $A \in \mathcal{B}(X)$, and let $M_{\Omega,X;\mathbb{P}}$ be the set of equivalence classes of $M_{\Omega,X}$ under the equivalence relation
$$ \dot{\mu}_1 \sim \dot{\mu}_2 \quad \Leftrightarrow \quad \textrm{for $\mathbb{P}$-a.a. $\omega \in \Omega$, $\ \dot{\mu}_1(\omega)=\dot{\mu}_2(\omega).$} $$
Now let $M_{\Omega \times X;\mathcal{F},\mathbb{P}}$ be the set of probability measures on the product space $(\Omega \times X, \mathcal{F} \otimes \mathcal{B}(X))$ with the property that $\mu(E \times X)=\mathbb{P}(E)$ for all $E \in \mathcal{F}$. Then $M_{\Omega,X;\mathbb{P}}$ and $M_{\Omega \times X;\mathcal{F},\mathbb{P}}$ are in exact one-to-one correspondence with each other, via the identification
$$ \mu(A) \ = \ \int_{\Omega \times X} \mathbf{1}_A(\omega,x) \, \dot{\mu}(\omega)(dx) \, \mathbb{P}(d\omega). $$
I don't know off hand any good textbook for the proof (the famous textbook on random dynamical systems by Ludwig Arnold gives the statement but I think it cites some other book - possibly not in English - for the proof). However, if you can't find the proof easily online, it is proved in my PhD thesis at https://spiral.imperial.ac.uk/handle/10044/1/39569 (Lemma 3.27 / Remark 3.28).
Corollary. For any $\dot{\mu} \in M_{\Omega,X}$ and any sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{F}$, there exists an element $\mathbb{E}[\dot{\mu}|\mathcal{G}]$ of $M_{\Omega,X}$ with the property that for all $A \in \mathcal{B}(X)$, the function $\,\omega \mapsto \mathbb{E}[\dot{\mu}|\mathcal{G}](\omega)(A)$ is a version of the conditional expectation $\mathbb{E}[\omega \mapsto \dot{\mu}(\omega)(A)|\mathcal{G}]$.
This is proved by taking the $\mu \in M_{\Omega \times X;\mathcal{F},\mathbb{P}}$ associated to $\dot{\mu}$, and then regarding $\mu$ as an element of $M_{\Omega \times X;\mathcal{G},\mathbb{P}|_\mathcal{G}}$ to go back in the other direction to get $\mathbb{E}[\dot{\mu}|\mathcal{G}]$. (Again, this can be found in Arnold's book.)
Thank you for your extensive and convincing analysis of the situation. I was aware of the fact, that ''nice structure of the state space $(X,\mathcal{B})$'' implies ''nice structure of the measure space $(M_X,\mathcal{M}_X)$'', where $\mathcal{M}_X$ denotes the evaluation $\sigma$-algebra. My question, however, aims at the converse direction (not rigorously, but by examples), that is, "too less structure of the state space$(X,\mathcal{B})$" leads to "too less structure of the measure space $(M_X,\mathcal{M}_X)$." Are there any examples? Thank you!
Well, certainly if $X$ is any non-metrisable then there does not exist a metrisable topology on $M_X$ for which the map $x\mapsto\delta_x$ is a topological embedding. So that's one obvious lack of nice structure. Now as for separability: if, say, $X$ is an uncountable set equipped with the discrete topology, then $\mathcal{B}(X)$ and hence $\mathcal{M}_X$ are not countably generated. (More generally, I think there's a result assuming the axiom of choice that the Borel $\sigma$-algebra of a non-separable metric space is never countably generated.
If $X$ is a non-Lusin separable metric space, then the topology of weak convergence on $M_X$ is not Lusin. I'm afraid I don't know of counterexamples to the disintegration theorem for non-Lusin separable metric spaces. By the way, non-separability of metric spaces causes all kinds of fundamental problems to working with probabilities in general: e.g. if I recall correctly, the diagonal in $X\times X$ cannot be $\mathcal{B}(X)\otimes\mathcal{B}(X)$-measurable, which is really bad! Likewise, I think the set of Dirac masses will not be $\mathcal{M}_X$-measurable.
Thank you for your hints and comments! I think your hint concerning the measurability of the set of Dirac measures is what I was looking for. (cf. my example below).
|
2025-03-21T14:48:29.916891
| 2020-02-21T20:33:16 |
353273
|
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|
Stack Exchange
|
Why is the RT invariant from $\mathcal Z(\mathcal C)$ the (norm) square of the one from $\mathcal C$?
The relationship between Turaev-Viro/state-sum invariants and Reshetikhin-Turaev/surgery invariants is roughly that
$$\tau_{TV, \mathcal C}(M) = |\tau_{RT, \mathcal C}(M)|^2.$$
Here $\mathcal C$ is a modular category.
More generally, we have
$$\tau_{TV, \mathcal C} = \tau_{RT, \mathcal C} \otimes \bar \tau_{RT, \mathcal C},$$
where $\bar \tau_{RT, \mathcal C}$ is the "mirror" theory where you take opposite orientations.
One way to prove this is via careful analysis of the state-sums and how they relate to surgery presentations of your manifold, as in Turaev's book.
On the other hand, if we instead start with a spherical fusion category $\mathcal C$ (weaker than being a modular category: in particular, there doesn't need to be a braiding on $\mathcal C$) is is known that
$$\tau_{TV, \mathcal C} = \tau_{RT, \mathcal Z(\mathcal C)}$$
where $\mathcal Z(\mathcal C)$ is the Drinfeld center. This is an equivalence of 3-2-1 TQFTs.
Presumably this means that if I have a modular category $\mathcal C$, then
$$\tau_{RT, \mathcal Z(\mathcal C)} = \tau_{RT, \mathcal C} \otimes \bar \tau_{RT, \mathcal C} $$
Is this true? Is there a way to see this equivalence without factoring through the TV theory? It would be enough for me to understand this just for links in $S^3$, not necessarily the full TQFT.
If $C$ is modular, in particular it is factorizable so that $Z(C)$ is braided equivalent to $C \boxtimes \bar C$ where $\boxtimes$ is an appropriate tensor product and $\bar C$ is $C$ with opposite tensor product and inverse braiding. I feel it shouldn't be hard to show in general that $\tau_{C \boxtimes \bar C}$ is $\tau_C \otimes \bar \tau_C$, and this is clearly true for links in $S^3$.
Why is $Z(C)$ braided equivalent to $C \boxtimes \bar C$? Do you know a reference that might discuss this in more detail?
This is one of the several equivalent definitions of modularity/factorizability, see e.g. https://arxiv.org/abs/1602.06534 for a proof in the more general non-necessarily semisimple case and references for the semisimple one. Note also that in the semisimple case the definition of $\boxtimes$ is much easier.
Or https://arxiv.org/abs/math/0111205 for an older school proof in the semisimple case.
|
2025-03-21T14:48:29.917079
| 2020-02-21T23:12:59 |
353277
|
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Stack Exchange
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Conformal maps between simply connected domains with piecewise real algebraic boundary
Between polygons in $\mathbb C\cup\{\infty\}$ (including the "single side polygons", hemispheres, disks) the Schwartz-Christoffel mappings give arguably explicit conformal maps. For polygons with few angles those are well known special functions -hypergeometric for triangles to halfplanes and elliptic integrals for rectangles. We also have domains with piecewise quadratic boundaries to which we can write explicit transformations. For instance sectors of disks map to disks via powers -removing the origin to make this conformal- and to halfplanes via Möbius transformations.
My question is whether we can write explicit transformations between domains piecewise bounded by higher degree real algebraic curves, like cubic or quartic, and a disk. There are numerical methods but can we use special functions?
This is related to calculating periods in the sense of Kontsevich-Zagier, though my motivation is rather low-level.
EDIT: I may consider as "explicit" integrals of algebraic functions. If you can give formulas in the form of "generalized Schwarz-Christoffel integrals" but for arbitrary polynomials defining piecewise the boundary that would already be satisfying.
I disagree that the Schwarz-Christoffel formula is "reasonably explicit", except in the case of triangle and rectangle, and very few other cases. Even less explicit it is for polygons whose sides are arcs of circles. For this case, the case of circular quadrilaterals has been intensively studied. In no way you can call this "explicit". The mapping is a ratio of two solutions of the Heun equation, and about solutions of this equation not so much is known.
I disagree that the Schwarz-Christoffel formula is "reasonably explicit", except in the case of triangle and rectangle, and very few other cases. The reason is that Schwarz-Christoffel formula for $n\geq 3$ contains unknown "accessory" parameters. Determination of these parameters requires inversion
of some rather complicated integrals.
Even less explicit it is for polygons whose sides are arcs of circles.
For this case, only circular triangles are reasonably well understood.
(See Klein's book Forlesungen uber hypergeometrische Funktion. In English,
Caratheodory, Function theory, vol. II).
The case of circular quadrilaterals has been intensively studied since
the second half of 19th century. In no way you can call the conformal map
on a generic circular quadrilateral "explicit". The mapping is a ratio of two solutions of the Heun equation, and about solutions of this equation not so much is known. Some people would call them "special functions", but they are not included in Whittaker Watson, except the special case of Lame equation. So one can say that
there is no explicit answer in any sense already for a generic circular quadrilateral. Some special quadrilaterals were subject of much research.
The literature about them is enormous, it goes under the names "Heun equation", (a special case is the Lame equation), "accessory parameters",
"Painleve VI",
and there is a lot of "physics" literature, old and modern, with keywords like "conformal blocks", etc.
Of course there are some very special cases when regions are bounded
by other algebraic curves, like ellipse, parabola, some cycloids or lemniscates. But these are very special cases.
A reasonable account of what one can do explicitly is: Werner von Koppenfels, and Friedmann Stallmann,
Praxis der konformen Abbildung,
Berlin, Springer-Verlag. (1959). It is somewhat out of date but not much.
You can glance at my own recent papers (all available on the arxiv) dedicated to some special cases of circular quadrilaterals, and even one on pentagons,
arXiv:1611.01356.
Thanks. Can you give me good references for mappings between interior of quartics, similar to the lemniscate, and a halfplane? At least a good reference for lemniscates?
Ok, I've had a look at your paper. I figured out that disks are mapped to lemniscate type quartics via $z\mapsto z^2$. I still have to think how far I can adapt Schwartz-Christoffel mappings. I guess I have enough intuition for my needs. Don't hesitate to add any comment. Thank you.
Another type of domains with algebraic boundary for which a conformal map is known, sort of explicitly is called "quadrature domains".
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353282
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Stack Exchange
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An "analytic continuation" of power series coefficients
Cauchy residue theorem tells us that for a function
$$f(z) = \sum_{k \in \mathbb{Z}} a(k) z^k,$$
the coefficient $a(k)$ can be extracted by an integral formula
$$a(k) = \frac{1}{2\pi i}\oint f(z) z^{-k-1},$$
with a contour around zero. Now, there is nothing to prevent us from thinking of $a\colon \mathbb{Z} \to \mathbb{C}$ as a function over the complex domain, defined by the above integral. In this way, we have naturally "continued" the function $a(k)$ over the integers to one over the complex numbers. Is there any meaning to this beyond just something amusing?
For example, consider $f(z)=\exp(z)$. In this case, $a(k)=1/k!$ which one would think would be continued to $1/\Gamma(1+k)$. But at least if you attempt to plot what happens above, this is not the case. We get a function with Bessel-like oscillations (blue) which nevertheless coincides with $1/\Gamma(1+k)$ (orange) on integral points!
One can experiment with other choices of $f(z)$. When $f(z)=(z+1)^n$, it seems like the continuation actually coincides with what one would naturally think (the Beta function). Which is actually quite strange. For example for $n=5$ and $f(z) = 1 + 5 z + 10 z^2 + 10 z^3 + 5 z^4 + z^5$, how does this computation "learn" that the sequence $1,5,10,10,5,1$ actually corresponds to the binomial coefficients and thus should be continued to the Beta function? If we perturb one of the coefficients, say change $5 z^4$ to $2 z^4$, the result would look like a "perturbed" Beta function (blue below vs the Beta function in orange).
Any explanation (or references in the literature) for what is going on here?
Just a remark. Your formula for $a(k)$ with non-integer $k$ is not actually uniquely defined. Because of the branch cuts in $z^{-k-1}$ for non-integer $k$, the value of the integral depends precisely on the shape of the contour (well, more precisely on the location of your chosen branch cut and where it intersects the contour). I noticed that you didn't specify the contour in your post. So, as you change the contour, your blue curve will change, except for its intersections with the orange curve.
The ML expression in my answer matches well the first plot you have from whatever app you were using to evaluate the integral. The integral method used by Euler to interpolate the factorials to the gamma function is unique for the right-half plane and is extended by Newton interpolation, Hadamard finite-part regularization, or the Hankel contour method. The RMT interp. is equivalent to Euler's. The binomial coefficients $\binom{\beta}{n}$ can be interpolated to $\binom{\beta}{\alpha}$ by sinc function/cardinal series interpolation for $Re(\beta) > -1$ since the BC is band-limited in $\alpha$.
The usual way to interpret the integral $$\frac{1}{2\pi i} \oint \exp(z) z^{-s} \, \mathrm{d}z$$ for nonintegral $s$ is to take a Hankel contour, and in this case the value is indeed $\frac{1}{\Gamma(s)}$ (see 5.9.2). Without knowing how you produced these plots, I'm not sure what else to say about your first example.
This is also related to Ramanujan's master theorem, which expresses this analytic continuation as a Mellin transform. Hardy gave some conditions under which this identity can be made rigorous.
Interesting, thanks. I just used a circular contour around the origin for the plots. Not sure why Hankel contour produces a different result, though especially for such a "clean" function as exp(z).
@MCH The exp(z) is fine, it's the $z^{-s}$ which gives you difficulties. The contour matters quite a bit. You can extend your circular contour to the Hankel contour by adding lines along the negative real axis so I suppose the function in your plot will be $1 / \Gamma(s)$ plus or minus some expression like $(e^{2\pi i s} - 1) \int_R^{\infty} e^{-t} t^s , \mathrm{d}t$ if the circle contour had radius $R$ (I haven't checked the details.)
Interesting. I'll check it now.
On RMT and the Mellin transform, see, e.g., https://mathoverflow.net/questions/79868/what-does-mellin-inversion-really-mean/79925#79925
The Hankel contour is one method. There are others, like the Pochhammer contour for the Euler beta function or the Hadamard finite-part-type of extensions. Choosing a branch cut is not too difficult a matter.
It is easy to see that for every convergent power series
$$f(z)=\sum_{n=0}^\infty a_nz^n$$
there exists an entire function of exponential type $F$ which interpolates
the coefficients: $F(n)=a_n$. This function is not unique.
But existence of such a function with some additional properties is related to analytic continuation properties of the power series.
Interpolating coefficients of a power series by an entire function of $n$ is a common powerful method of the study of singularities of power series. There is a close relation between the growth of this entire function and analytic continuation of the power series. See, for example,
L. Bieberbach, Analytische Fortsetzung, Springer 1955,
or V. Bernstein, Lecons sur les progres recents de la theorie de series de Dirichlet, Paris, 1933.
Sorry, I do not know any books in English, but you can look at this paper
arXiv:0709.2360, Theorem C on p. 7 for a typical result.
Another nice result is the theorem of Leau:
A convergent power series
$$\sum_0^\infty a_nz^n$$
has an analytic continuation to $\overline{C}\backslash \{1\}$ if and only if
$a_n=f(n)$ where $f$ is an entire function of minimal exponential type.
And there are many theorems in between.
Where does this "theorem of Leau" come from?
It comes from the paper of Leopold Leau, Journ. de Math. (5) 5, 365-425 (1899).
As pointed out in comments, your extension of $a(k)$ depends on the choice of branch used for $z^{-k-1}$. All such extensions, though, are precisely the band-limited extensions of $a(k)$ given by the Nyquist-Shannon sampling theorem. For instance, if one uses the standard branch cut on the negative real axis, then the extension
$$ a(x) = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(e^{i\theta}) e^{-ix\theta}\ d\theta$$
is the unique extension of
$$ a(k) = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(e^{i\theta}) e^{-ik\theta}\ d\theta$$
from the integers to the reals which is band-limited to the spectral interval $[-1/2,1/2]$ (or $[-\pi,\pi]$, depending on one's choices of convention for Fourier transform). The extension can be given explicitly by the Whittaker-Shannon interpolation formula
$$ a(x) = \sum_{k \in {\bf Z}} a(k) \mathrm{sinc}(\pi(x-k))$$
where $\mathrm{sinc}(x) = \frac{\sin(x)}{x}$ is the sinc function (with convention $\mathrm{sinc}(0)=1$). Other choices of branch cut lead to slightly different formulae of this type.
The reason why this procedure gives the "wrong" extension for the reciprocal Gamma function $1/\Gamma(k+1)$ is simply that this function is not band-limited. From the Hankel contour formula mentioned in another answer one instead has
$$ \frac{1}{\Gamma(k+1)} = \frac{1}{2\pi i} \oint e^z z^{-k-1}\ dz + \frac{\sin(\pi k)}{\pi} \int_1^\infty e^{-t} t^{-k-1}\ dt$$
(I have not checked for sign errors etc.); the second term is not band-limited and is causing the discrepancy between your band-limited interpolation and the interpolation you were hoping for.
This argument can be couched in terms of Mellin transform/Newton interpolation versus Fourier transform/sinc function (cardinal series) interpolation https://mathoverflow.net/questions/192146/newton-series-and-fourier-transform-is-there-an-analogy?noredirect=1&lq=1
While reordering the database of my library, I found a reference to paper [1], which I think is somewhat pertaining to the question. The Authors, in order to prove a result on the localization of singularities of lacunary power series on the boundary of their circle of convergence, establish a necessary and sufficient condition for the analytic continuation across boundary arcs, which involves exactly the possibility of "continuing" the function $a:\Bbb N\to\Bbb C$ (in our notation) to the whole complex plane. Precisely, they prove the following result
([1], theorem 2, §2.2, pp. 564-567) If $\sigma \in [0,\pi )$, then the open arc
$$
\gamma_\sigma\triangleq \{e^{i \theta}: \sigma < \theta < 2\pi - \sigma\}
$$ is an arc for regularity of the power series $f$ (i.e. it can be analytically continued across it) if and only if there exists an entire function of exponential type $\varphi$ satisfying the following conditions
$\varphi (n) = a(n)$ for $n \in\mathbb N$,
$h_\varphi (0) = 0$ and
$$
\limsup_{\theta \to 0} \frac{h_\varphi (\theta )}{| \theta |} \leq \sigma
$$
where $h_\varphi (\theta )$ is the indicator function of $\varphi$, i.e.
$$
h_\varphi (\theta )\triangleq\limsup_{r \to \infty}\big (r^{-1} \log | \varphi (r e^{i \theta})| \big)
$$
The proof is constructive: the entire function is constructed in the form
$$
\varphi(\zeta)=\frac{1}{2\pi i}\oint\limits_{\Gamma_{\gamma_\sigma}} f(z) z^{-\zeta-1}\mathrm{d} z,
$$
where $\Gamma_{\gamma_\sigma}$ is a properly constructed simple closed path (and this possibly accounts for the nice observation by Igor Khavine).
Final note
Theorem 2 of [1] above states that the possibility of representing the coefficients of $f$ as the evaluation over the integers of an exponential entire function belonging to a particular class is equivalent to its analytic continuablity across an open arc of the circumference of its convergence disk. However, if $f$ is non continuable across the boundaries of convergence disk (i.e. its circumference is a "cut" for $f$) the above theorem is not applicable: I'd like to know if, in such case, the Fourier integral approach described by prof. Tao in his answer could be used to say something on the structure of this kind of entire functions.
Reference
[1] Norair Arakelian and Wolfgang Luh and Jürgen Müller, "On the localization of singularities of lacunary power series", (English)
Complex Variables and Elliptic Equations 52, No. 7, 561-573 (2007), MR2340942, Zbl 1123.30001.
Edit (1/21/21): (Start)
For your first example, a classic method of interpolation is related to fractional calculus.
$$k!\; a(k) = k! \; \oint_{|z|=r} \frac{e^z}{z^{k+1}} \; dz = e^{-1}k! \; \oint_{{|z|=r}} \frac{e^{z+1}}{z^{k+1}} \; dz $$
$$= e^{-1}k! \; \oint_{|z-1|=1} \frac{e^{z}}{(z-1)^{k+1}} \; dz =e^{-1} D^k_{z=1} e^z.$$
Interpolating using a standard fractional integroderivative, of which there are several reps,
$$\lambda! \; a(\lambda) = \; e^{-1} D_{z=1}^{\lambda} \; e^z = e^{-1} \; \sum_{n \ge 0} \frac{z^{n-\lambda}}{(n-\lambda)!}\; |_{z=1}$$
$$ = e^{-z} \; \sum_{n \ge 0} \frac{z^{n-\lambda}}{(n-\lambda)!}\; |_{z=1} = e^{-z} z^{-\lambda}\; E_{1,-\lambda}(z) \; |_{z=1},$$
where $E_{\alpha,\beta}(z)$ is the Mittag-Leffler function (general definition in Wikipedia, MathWorld; some history and applications), encountered very early on by anyone exploring fractional calculus.
This method of interpolation gives the entire function
$$ a(\lambda) = e^{-1} \; E_{1,-\lambda}(1) \frac{1}{\lambda!} = e^{-1} \; \sum_{n \ge 0} \frac{1}{(n-\lambda)!} \; \frac{1}{\lambda!}, $$
which matches the OP's first graph and extends it to any real or complex argument, giving, of course, $a(k) = 1/k!$ for $k=0,1,2, ...$.
Similarly, the fractional calculus can be applied to usefully interpolate coefficients generated by $ D_z^n \; f(z)$ to those of $D_z^{\lambda} f(z)$, with care taken with defining branch cuts and contours of integration for the complex function for the point of evaluation. The fundamental example is Euler's integral rep of the beta function as depicted below. Another is the interpolation of the associated Laguerre polynomials to the confluent hypergeometric functions and, therefore, interpolations of the associated coefficients. (See links below.)
Other productive methods of interpolation--all can be related to fractional calculus--are sketched below.
(End)
Edit 1/23/21: (Start)
The interpolation you desire is obtained by applying Ramanujan's favorite Master Formula, a.k.a. Mellin transform interpolation, as discussed below and illustrated in four other MO-Qs--Q1, Q2, Q3, and Q4. This, naturally, amounts to replacing $k$ by $s$ only outside your Cauchy integral rather than within for the reason Terry Tao notes, which is equivalent to noting the Cauchy contour integration is not equal to the Mellin transform integration.
(End)
The Mellin transform pair allows for interpolation of the coefficients of generating functions, often directly connected to sinc and/or Newton interpolation. Basically, the following is a sketch of the analytics of Ramanujan's Master Formula/Theorem, which he so profoundly and elegantly wielded.
Here $(a.)^n := a_n$ in the Taylor series $e^{a.x}$ is interpolated as $a_{-s}$ via the Mellin transform of the Taylor series $f(x) = e^{-a.x}$.
First consider the normalized Mellin transform and its inverse
$$F(s) = MT[f(x)] = \int_{0}^{\infty} f(x) \; \frac{x^{s-1}}{(s-1)!} \; dx$$
$$f(x) = MT^{-1}[F(s)] = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} F(s) \frac{x^{-s}}{(-s)!} \; ds .$$
Then the RMT holds for a class of functions such that the simple poles of the inverse sine in the inverse Mellin transform give
$$f(x) = e^{-a.x} = \sum_{n \geq 0} \frac{(-a.x)^n}{n!} = \sum_{n=0} a_n \frac{(-x)^n}{n!} =
\sum_{n=0} F(-n) \frac{(-x)^n}{n!} \; ,$$
that is, such that we may close the complex contour to the left (e.g., in the sense of the limit of a semicircle with its radius expanding to infinity) for $0 < \sigma < 1$ and $0 < x < 1$ when $F(s)$ has no singularities/poles within the contour. This rep allows an extension of the RMT (and the Mellin transform) to cases in which poles are present in $F(s)$ and to other ranges of $x$.
Also note (see, e.g., Gelfand and Shilov's "Generalized Functions") the relation
$$D_x^{m+n+1} \; H(x) \frac{x^m}{m!} = H(x) \frac{x^{-n-1}}{(-n-1)!} = \delta^{(n)}(x),$$
reflected in the two (of several) reps of the fractional differintegro op equivalent under analytic continuation
$$\frac{x^{\alpha-\beta}}{(\alpha-\beta)!} = \frac{d^{\beta}}{dx^\beta}\frac{x^{\alpha}}{\alpha!}=\int_{0}^{x}\frac{z^{\alpha}}{\alpha!}\frac{(x-z)^{-\beta-1}}{(-\beta-1)!} dz = \frac{1}{2\pi i} \oint_{|z-x|=|x|}\frac{z^{\alpha}}{\alpha!}\frac{\beta!}{(z-x)^{\beta+1}}dz ,$$
with $H(x)$ the Heaviside step function and $\delta(x)$, the Diraac delta. (This is equivalent through binomial expansion to a sinc function/cardinal series interpolation of the binomial coefficients $\binom{s}{\alpha} = \sum_{n \geq 0} \binom{s}{n} \frac{\sin(\pi(\alpha-n))}{\pi(\alpha-n)}=\sum_{n \geq 0} \binom{s}{n} \binom{0}{\alpha-n} $, an instance also of the Chu-Vandermonde identity. The contour integral for this beta function integral is easily transformed into a bandlimited Fourier transform. See, e.g., the relation to Newton interpolation in the MSE_Q "Why is the Euler gamma function the best extension of the factorial function to the reals?" For more detail and connections, see my post "Fractional calculus and interpolation of generalized binomial coefficients." The integral reps of the confluent (see this MO-Q) and non-degenerate hypergeometric functions can be expressed effectually as this or easily related differintegral ops acting on simple functions. Years ago I found the Danish masters Niels Nielsen and Niels Norlund to be most informative on contour integrals in interpolation. There is also a connection to Riemann surfaces via Pochhammer's contour for the beta function integral.)
So, under the conditions above,
$$F(-n) = \int_{0}^{\infty} f(x) \; \frac{x^{-n-1}}{(-n-1)!} \; dx = \int_{0}^{\infty} e^{-a. x} \; \delta^{(n)}(x) \; dx = a_n,$$
and this suggests the analytic continuation and relation to umbral calculus
$$F(s) = \int_{0}^{\infty} f(x) \; \frac{x^{s-1}}{(s-1)!} \; dx = \int_{0}^{\infty} e^{-a.x} \; \frac{x^{s-1}}{(s-1)!} \; dx = (a.)^{-s} = a_{-s}.$$
The iconic guiding example is the Euler gamma function integral rep with $(a.)^n = a_n = c^n$
$$ (a.)^{-s} = a_{-s} = c^{-s} = F(s) = MT[f(x)= e^{-c\; x}] = \int_{0}^{\infty} e^{-c \; x} \; \frac{x^{s-1}}{(s-1)!} \; dx = \frac{1}{c^{s}},$$
giving the interpolation of the coefficients of the Taylor series of $e^{cx}$, i.e., $a_n = c^n$, as $a_{-s}=c^{-s}=F(s)$.
Another useful example, which vividly illustrates the relation to the Appell Sheffer sequences of umbral calculus (of which the $x^n$ with e.g.f. $e^{x}$ is the basic example), is the integral rep for (what I call) the Bernoulli function, simply related to the Hurwitz zeta function and generalizing the Bernoulli polynomials,
$$ B_{-s}(z) = (B.(z))^{-s} = \int_{0}^{\infty} e^{-B.(z)t} \; \frac{t^{s-1}}{(s-1)!} \; dt $$
$$ = \int_{0}^{\infty} \frac{-t}{e^{-t}-1} \; e^{-zt} \frac{t^{s-1}}{(s-1)!} \; dt = s \; \zeta(s+1,z)$$
where the e.g.f. for the Bernoulli polynomials with $(b.)^n = b_n$ the Bernoulli numbers is
$$e^{B.(x)t} = e^{(b.+x)t} = e^{b.t} e^{xt} = \frac{t}{e^t-1} \; e^{xt}.$$
Note that
$$B_n(z) = -n \; \zeta(1-n,z),$$
$$B_n(1) = -n \; \zeta(1-n,1) =-n \; \zeta(1-n) (Riemann) = (-1)^n B_n(0) = (-1)^n b_n.$$
Through this characterization, it is not too difficult to show that the Bernoulli function inherits all the elegant properties of a regular Appell sequence, such as $D_z \; B_{s}(z) = s \; B_{s-1}(z)$.
Hankel contour deformations, Hadamard finite part regularization of the Mellin transform guided by the inverse Mellin transform, the Mellin-Barnes contour integral, and other methods of analytic continuation can be used to extend the range of $s$ and other parameters, as have been done for the integral reps of the Euler gamma and beta functions and Riemann and other zeta functions and their generalizations. In addition, interchanging summation and integration often gives rise to useful asymptotic expansions of functions a la Borel, Heaviside, Hardy, and Poincare.
Riemann knew all this stuff. Ramanujan intuited it. Hardy formalized it. (I stumbled across it on a journey starting from the ladder ops of QM and a brief comment by my old math prof Stallybrass about the sequence $D^{m+n} H(x) \frac{x^m}{m!}$ in his integral transforms class an eon ago.)
For application to defining fractional powers of operators, see my answer and comments therein to the MO-Q "What does the inverse Mellin transform really mean?" and several of my blog posts, such as "The Creation / Raising Operators for Appell Sequences."
For the history of sinc function/cardinal series interpolation, see the MO-Q "History of the sampling theorem" https://mathoverflow.net/questions/97180/history-of-the-sampling-theorem/97202#97202
An example of Hadamard-finite-part continuation: MSE-Q "Domain of the gamma function" (https://math.stackexchange.com/questions/13956/domain-of-the-gamma-function/132727#132727). Example of the Hankel-contour continuation: MO-Q "How does one motivate the analytic continuation of the Riemann zeta function?" (https://mathoverflow.net/questions/58004/how-does-one-motivate-the-analytic-continuation-of-the-riemann-zeta-function/97401#97401).
A little more than just amusing. This type of analytic continuation via the Mellin transform was used by Hawking in "Zeta function regularization of path integrals in curved spacetime." You'll find the Hurwitz zeta function in "Zeta-function regularization is uniquely-defined and well" by Elizalde to calculate the Casimir effect for certain closed strings and in many other articles dating from Schwinger in calculations in quantum field theory, all related to analytic continuation of general zeta functions defined through the Mellin transform, i.e., zeta function regularization, I believe.
On some relations between the Mellin transform and Newton interpolation and between the Fourier transform and sinc fct/cardinal series interpolation, see https://mathoverflow.net/questions/192146/newton-series-and-fourier-transform-is-there-an-analogy?noredirect=1&lq=1
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353286
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Stack Exchange
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$\Sigma^2_1$ and the Continuum Hypothesis
This is a follow up to Will Brian's answer to this recent question. In particular, quoting Brian:
"In fact, Paul Larson has pointed out to me that the statement "$\phi$ and $\phi^{-1}$ are conjugate" is a now very rare example of a $\Sigma^2_1$ statement about the real line whose status we do not know under ZFC+CH."
(A $\Sigma^2_1$ statement is one that can be expressed in the form $\exists A\subseteq\mathbb{R}\psi(A,r)$ where $r$ is a fixed real parameter and $\psi$ is a formula whose quantifiers range only over $\mathbb{R}$.)
How rare is "very rare"? Are there any other known examples of $\Sigma^2_1$ statements concerning the real line whose status under CH (+ large cardinals) is unknown? What are some other examples?
This should say ZFC+CH+large cardinals, or we get silly projective statements.
Thanks @AndrésE.Caicedo!
Well another example, though of the same kind, is the following: let $(p_n)$ and $(p'_n)$ be the increasing sequence of primes $=1$ mod $4$, resp. $=3$ mod $4$ (this precise choice doesn't matter), let $c$ be a permutation with one $p_n$-cycle for each $n$ and no other cycle, and similarly define $c'$. Are $c$ and $c'$ conjugate as self-homeomorphisms of the Stone-Cech remainder? I just guess it's unknown, I'm not sure anyone else ever thought about it.
As an outsider, I don't guess what's the meaning of "under CH+ large cardinals". Does it mean "under CH, assuming the consistency of existence of some kind (what kind?) of large cardinal"? or *"under CH + existence of some kind (what kind??) of large cardinal"? is the [consistency of the?] existence of a strongly inaccessible cardinal enough to avoid the pathologies mentioned by Andrés? By the way I'd also like to see some "typical" illustrating examples of theorems of ZFC or ZFC+CH reducing to a $\Sigma_1^2$ statement.
This is all motivated by a result of Woodin that $\Sigma^2_1$ statements are absolute for set forcing between models of ZFC+CH+"enough large cardinals". So if there are enough large cardinals around, there is a kind of "canonical" theory of ZFC+CH at the level of $\Sigma^2_1$. The question of whether $\phi$ and $\phi^{-1}$ are conjugate has the right complexity, but it is unknown what the answer is in the canonical theory. (Also, Andres is much more of an expert on this than I am, so he may be able to say more intelligent things here.)
And indeed, he has. See
https://mathoverflow.net/questions/108197/sets-of-reals-and-absoluteness/108200#108200
@AndrésE.Caicedo: Thanks for pointing that out. As you could probably guess, the error is mine, not Paul's. Todd, one answer to your question, which I also got from Paul Larson, is "there is a self-homeomorphism $\psi$ of $\omega^*$ that commutes with the shift map $\phi$ but is not a power of it." Like the example in YCor's comment, I'm not sure this should really qualify as an answer, because it seems a bit too close to the statement we started with. It would be nice to see an example that doesn't just ask about pathological automorphisms of $\mathcal P(\omega)/\mathrm{fin}$.
|
2025-03-21T14:48:29.919302
| 2020-02-22T02:52:41 |
353287
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Emil Jeřábek",
"Fedor Petrov",
"Louis Deaett",
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}
|
Stack Exchange
|
Smallest $S\subset \mathbb C$ on which no degree $k$ polynomial always vanishes?
Say $p$ is a polynomial of degree $k$ in $\mathbb C[x]$. Then $p$ can have at most $k$ distinct roots. A somewhat obtuse way to state that is to say that among any set of $k+1$ distinct complex numbers, there must exist a value $a$ for which $p(a)\neq 0$.
The question here has to do with generalizing the above fact to multivariate polynomials.
Given some $p \in \mathbb C[x_1,\ldots,x_n]$, it turns out that whenever we have enough different values to choose from, we can always pick from these some $a_1,\ldots,a_n$ so as to give $p(a_1,\ldots,a_n)\neq 0$. Moreover, how many is "enough" is a function just of $n$ and the degree of $p$.
I want to emphasize that, in the above and in what follows, the $a_1,\ldots,a_n$ so chosen are distinct.
To make things more precise and specific, we have the following.
Theorem. Let $p \in \mathbb C[x_1,\ldots,x_n]$ be a polynomial of total degree $k$. Then any set of $n+k$ distinct complex numbers contains a subset of $n$ that can be assigned to the variables $x_1,\ldots,x_n$ so as to give a nonzero value for $p$. (Again, we are talking about choosing from the set $n$ distinct values that we assign to $x_1,\ldots,x_n$.)
The proof of the above is a not-too-difficult exercise.
But the question is:
Question. In the above theorem, can the $n+k$ be replaced with some other function of $n$ and $k$ so as to give a better upper bound?
In fact, let's define $M(n,k)$ to be the smallest positive integer such that for every $p\in \mathbb C[x_1,\ldots,x_n]$ of total degree $k$, every set $S$ of at least $M(n,k)$ distinct complex numbers contains some subset $\{a_1,\ldots,a_n\}$ (of cardinality $n$, so that these are distinct values) such that $p(a_1,\ldots,a_n)\neq 0$.
The theorem above states that $M(n,k) \le n+k$, but I don't even see how to meet that bound for $n=k=2$.
This is probably just another way to present Fedor Petrov's solution: Expand
$$\frac{(1-t \alpha_1) (1-t \alpha_2) \cdots (1-t \alpha_{n+k-1})}{(1-t \beta_1)(1 - t \beta_2) \cdots (1-t \beta_n)}$$
as a formal power series in $t$. The coefficient of $t^k$ is a degree $k$ polynomial in the $\beta$'s, which vanishes whenever $\{ \beta_1, \beta_2, \ldots, \beta_n \}$ is a $n$-element subset of the $\alpha$'s.
Three does not suffice for $n=k=2.$ Consider $x^2+xy+y^2-1$ with the set $\{-1,0,1\}.$
Oh, nice. Now I'm not sure whether I suspect the bound can be met in general.
moreover, $n+1$ do not suffice for $k=2$ and any set $A$ of size $n+1$. Consider the polynomial $x_1^2+\ldots+x_n^2+(x_1+\ldots+x_n-s)^2-t$, where $s=\sum_{a\in A} s$, $t=\sum_{a\in A} a^2$.
... and the same trick works for any $k$
No. Moreover, for any set $A\subset \mathbb{C}$, $|A|=n+k-1$, you may find a not identically zero polynomial $p_A$ of degree at most $k$ such that $p_A(x_1,\ldots,x_n)=0$ for all distinct $x_i\in A$. For $n=1$ this is already mentioned in OP, so further I suppose that $n>1$.
For $k-1$ variables $t_1,\ldots,t_{k-1}$ denote $p_i=\sum_{j=1}^{k-1} t_j^i$. We may express $p_k$ as a polynomial in $p_1,\ldots,p_{k-1}$. This gives us a polynomial $h$ such that $h(p_k,p_{k-1},\ldots,p_1)\equiv 0$, and $h(z_k,\ldots,z_1)$ is weighted homogeneous with weighted degree $k$: for any monomial $\prod z_i^{c_i}$ in $h$ we have $\sum ic_i=k$. Next, consider $k$-th variable $t_k$ and the polynomial $$F(t_1,\ldots,t_k):=h\left(\sum_{j=1}^{k} t_j^k,\sum_{j=1}^{k} t_j^{k-1},\ldots,\sum_{j=1}^{k} t_j\right).$$
$F$ is symmetric, not identically zero (since $h$ is not identically zero and the guys $\sum_{j=1}^{k} t_j^i$ which we substitute to $h$ may take any complex values), is homogeneous of degree $k$. Also $F$ takes zero value when one of $t_i$'s is zero. Therefore $F$ is divisible by $t_1\ldots t_k$, and since $\deg F=k$ we get $F=c\,t_1\ldots t_k$ for certain non-zero constant $c$.
Now denote $\alpha_i=\sum_{a\in A} a^i$ and take the polynomial
$$
p_A(x_1,\ldots,x_n)=h\left(\alpha_k-\sum x_i^k,\alpha_{k-1}-\sum x_i^{k-1},\ldots,\alpha_1-\sum x_i\right).
$$
From the above properties of $h$ we get $\deg p_A\leqslant k$ and $p_A(x_1,\ldots,x_n)=0$ whenever $x_1,\ldots,x_n$ are distinct elements of $A$. It remains to show that $p_A$ is not identically zero. For this aim we choose $k$ non-zero elements $a_1,\ldots,a_k\in A$, denote other elements of $A$ by $a_{k+1},\ldots,a_{k+n-1}$ and substitute $x_i=a_{k+i}$ for $i=1,\ldots,n$ and $x_n=0$ to $p_A$. This value of $p_A$ is nothing but $F(a_1,\ldots,a_k)\ne 0$.
Why is the total degree of $p_A$ $k$?
Because $h(p_k,p_{k-1},\ldots,p_1)$ is homogeneous of degree $k$ (where $p_i=\sum_{i=1}^{k-1} t_i^k$), thus degree of $p_A$ is at most $k$. If it is less than $k$, multiply it by something. If it is identical 0, well, we need some additional argument. But at least it is not identical zero for some some sets $A$, that is ok for OP.
I see. Would you care to explain why $h$ with such degree bounds exists? I understand that $p_1,\dots,p_k$ are algebraically dependent, but this seems to be a much stronger property.
@EmilJeřábek3.0 it is some basics of symmetric polynomials: any symmetric polynomial may be expressed as a polynomial in $p_1,\ldots,p_{k-1}$, in particular $p_k$.
Yes. But why it can be done with such tight constraints on the monomials?
Well, if $p_k=H(p_1,\ldots,p_{k-1})$ for some polynomial $H$, then any monomial $p_1^{c_1}\ldots p_{k-1}^{c_{k-1}}$ in $H$ is homogeneous of degree $\sum ip_i$ as a polynomial in $t_1,\ldots,t_{k-1}$. So if we take homogeneous components of degree $k$ in both parts of the equality $p_k=H(p_1,\ldots,p_{k-1})$, all such wrong monomials disappear from $H$.
Oh I see, thank you.
|
2025-03-21T14:48:29.919670
| 2020-02-22T02:58:15 |
353288
|
{
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|
Stack Exchange
|
Higher order functional derivatives
Let $E, F$ be Banach spaces. A continuous bilinear functional ${\langle \cdot\,, \cdot \rangle }: E \times F \to \mathbb{R}$ is called $E$-non-degenerate if $\langle x,y\rangle = 0$ for all $y \in F$ implies $x=0$ (Similarly for $F$-non-degenerate). Equivalently, the two maps of $E$ to $F^{*}$ and $F$ to $E^{*}$ defined by $x \mapsto \langle x, \cdot \rangle$ and $y \mapsto \langle \cdot\,, y\rangle$, respectivelly, are one-to-one. If they are isomorphisms (*), $\langle \cdot\,, \cdot \rangle$ is called $E$ or $F$-strongly non-degenerate. We say that $E$ and $F$ are in duality if there is a non-degenerate bilinear functional $\langle \cdot\,, \cdot \rangle: E\times F \to \mathbb{R}$, also called a pairing of $E$ with $F$. If the functional is strongly non-degenerate, we say the duality is strong.
Consider the following definition.
Definition: Let $E$ and $F$ be normed spaces and $\langle \cdot, \cdot \rangle$ a $E$-non-degenerate pairing. Let $f: F \to \mathbb{R}$ be Fréchet differentiable at the point $\alpha \in F$ (denote this derivative as $Df(\alpha)$). The functional derivative $\delta f/\delta \alpha$ of $f$ with respect to $\alpha$ is the unique element in $E$, if it exists, such that:
\begin{eqnarray}
Df(\alpha)(\gamma) = \left\langle \frac{\delta f}{\delta \alpha}, \gamma\right\rangle\quad\forall\gamma \in F. \tag{1}\label{1}
\end{eqnarray}
Now, I'd like to know how to define higher order derivatives of functional derivatives. In other words, suppose the Fréchet derivative of $f$ at $\alpha$, $Df(\alpha)$ is Fréchet differentiable at $\beta\in F$, is it possible to define $\dfrac{\delta^{2}f}{\delta \beta\delta\alpha}$?
Now, I'd like to know how to define higher order derivatives of functional derivatives. In other words, suppose the Fréchet derivative of $f$ at $\alpha$, $Df(\alpha)$ is Fréchet differentiable at $\beta\in F$, is it possible to define $$
\dfrac{\delta^{2}f}{\delta \beta\delta\alpha}?
$$
Yes, it is possible to define higher order Fréchet derivatives directly as bilinear functionals in $\mathscr{L}_2(F,\Bbb R)$ (the space of bounded bilinear functionals from $F$ to $\Bbb R$). This is shown by Ambrosetti and Prodi ([1], pp. 23-29), for example, and while leaving to them the details, we can simply say that $f:F\to\Bbb R$ is twice Fréchet differentiable at $\alpha\in F$ iff
$F$ is one time Fréchet differentiable and
$Df[\alpha+\beta](\gamma)-Df[\alpha](\gamma)= \mathfrak{B}[\alpha](\beta,\gamma) + o(\beta)$
where $o(\beta)/\Vert\beta\Vert_F\to 0$ as $\Vert\beta\Vert_F\to 0$ and $\mathfrak{B}\in\mathscr{L}\big(F,\mathscr{L}(F,\Bbb R)\big)\simeq\mathscr{L}_2(F,\Bbb R)$ (again, the proof of this fact is found in [1] §1.3, p. 23, where it is precisely show that the isomorphism between these two spaces is actually an isometry).
This can bee seen also by using the definition of Fréchet derivative you give, but it seems to me that, when dealing with functional derivatives of order greater than one, it overshadows the intrinsic clarity and familiarity of the concept. Precisely, by using the standard definition of Fréchet differentiablity (as given for example in [1] §1.1, p. 9, definition 1.1) and \eqref{1} we get
$$
\begin{split}
Df[\alpha+\varepsilon\beta](\gamma)-Df[\alpha](\gamma) &= \left\langle \frac{\delta f}{\delta \alpha}(\alpha+\beta),\gamma\right\rangle -\left\langle \frac{\delta f}{\delta \alpha} (\alpha),\gamma\right\rangle\\
& = \left\langle\frac{\delta f}{\delta \alpha}(\alpha+\beta)- \frac{\delta f}{\delta \alpha} (\alpha),\gamma\right\rangle\\
&=\left\langle\mathfrak{L}[\alpha](\beta)+ o(\beta),\gamma\right\rangle
\end{split}
$$
where $\mathfrak{L}\in\mathscr{L}(F,E)$ is a linear bounded operator.
Bibliography
[1] Ambrosetti, Antonio; Prodi, Giovanni, A primer of nonlinear analysis, Cambridge Studies in Advanced Mathematics, 34. Cambridge: Cambridge University Press, pp. viii+171 (1993), ISBN: 0-521-37390-5, MR1225101, ZBL0781.47046.
I would even say: there is but one notion, the Fréchet differential; which allow a unified treatment of the theory. Then, in each individual space of the luxuriant family of Banach spaces we have special representation theorems for duals, k-linear forms, and linear operators, so that the only abstract notion takes plenty of different forms, and "functional derivative" is just one of these.
Thanks for the answer and comment guys! I didn't read it yet (gonna do it soon) but have you seen my latest question? It's an extension of this one and you might be able to contribute a lot! https://mathoverflow.net/questions/355279/functional-derivatives-on-banach-spaces?noredirect=1#comment891495_355279
@IamWill, you're welcome. I saw your latest question but I wanted to answer this one first because the definition you give here (which I learned from you that it is due to Abraham, Marsdeen and Ratiu) fooled me a little bit. Then I realized that you only asked how to define higher order Fréchet derivatives by using it: now I hope to have shown that this definition is not the most manageable one when having to deal with higher order differentiability of functionals, but nevertheless it gives the standard result.
@DanieleTampieri I just read you answer and it helps me a lot! Let me just clarify one thing: your $\mathcal{L}(\alpha)$ is a linear operator which is the first order derivative of $f$ at $\alpha$?
@IamWill yes, it is a bounded linear operator. When a Fréchet derivative is involved, in general $n$-linear bounded (and thus continuous) operators are involved: I'll add an explanation to the answer above.
Yes, forgot the bounded condition. It makes very much sense to me but how can I assure I can write $Df\alpha+\beta-Df\alpha$ as $\langle \mathcal{L}\alpha,\gamma\rangle$? I got confused at this point. I mean, the factos $\delta f/\delta \alpha$ are not derivatives per se, they're elements of $F$. Did you use any representation theorem or something or it is easier than I think?
@IamWill, yes, it is easier than you think: It is simply a consequence of the definition you used and of the assumption of Fréchet differentiability of the Fréchet derivative of $F$. I'll add something on this to my answer above.
@DanieleTampieri thank you so much! It helps a lot!
|
2025-03-21T14:48:29.920043
| 2020-02-22T06:40:14 |
353294
|
{
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"Daniele Tampieri",
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"tim"
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}
|
Stack Exchange
|
Accuracy of solutions of system of multivariable polynomials
Theory of Gröbner basis has significant implications in many branches of Mathematics and Science.
In particular, Gröbner basis are used to solve systems of multivariable polynomials that appears in different branches of Mathematics or Science.
For a solvable system of multivariable polynomials, there exists a univariant polynomials (i.e., polynomials in one unknown $x_1$) in the Gröbner basis. The first step to solve the system is to solve that univariant polynomials.
Due to the fact that degree $\ge 5$ univariant polynomials cannot be solved in terms of radicals, the computed roots of that univariant polynomials must be an approximation of the actual roots. Then the approximated roots are substituted back to the system to get another univariant polynomial in some other unknown $x_2$. In other words, we will get an approximation of actual roots of the approximation of the univariant polynomial in $x_2$. The process repeats until we get all the approximations of the solutions of the system.
How can we know the approximations of solutions are small enough to make mathematical/ scientific conclusion? For example, can we get an approximation of the solution such that the error between the approximation and the actual solution are $<10^{-16}$?
Currently, I am using wxmaxima to compute the Gröbner basis.
For some cases, I know $x_i=\dfrac{\sqrt{3}}{2}$ for some $i$, but the wxmaxima gives me an approximation with error of order $10^{-7}$.
How can I improve the approximation so that the error is $<10^{-16}$ and how can I know the error for other $x_j$ in general?
There are algorithms to find the roots of univariate polynomials with guaranteed precision, see the MPSolve software: https://numpi.dm.unipi.it/software/mpsolve This does not answer your question though, since you also want to compute $x_2,x_3 \ldots$. A naïve approach is to compute the univariate polynomials wrt each $x_1,x_2,x_3 \ldots$, compute the approximate roots of those, and then decide which tuples $(x_1,\ldots,x_n)$ satisfy your system of equations.
Very interesting question: a subtly related one was posted as a comment to this answer last summer by @semiclassical. He asked how it is possible to recover the exact radical solutions to a given univariate polynomial equation from the general algorithm, and this in turn is equivalent to ask how to recover a well defined and fast converging algorithm to a given root from a general one.
The Arb library in Sage can compute roots of approximate polynomials (that is, polynomials whose coefficients are complex balls). The output should be balls guaranteed to contain the roots. So you could apply root-finding to your approximate polynomial in $x_2$, and so on. See http://arblib.org/acb_poly.html#root-finding for more information.
The numerical accuracy of symbolic elimination methods generally doesn't scale well with the number of variables or degrees. See "Numerical instability of resultant methods" by Noferini and Townsend for a striking result along these lines. If you want control over the accuracy, you are probably better off using a global method for multivariate solving (eg. homotopy continuation if you want all solutions) and polishing roots with Newton. There are various theorems about how to check you are with $\epsilon $ of a true solution by looking at the Newton step and related quantities.
|
2025-03-21T14:48:29.920277
| 2020-02-22T09:06:27 |
353297
|
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|
Stack Exchange
|
Fractional Sobolev norm of characteristic function of an interval?
Is there an explicit expression giving a fractional Sobolev norm of the characteristic function of some interval $I=[a,b)$?
I believe it is true that $\chi_{I} \in W^{s,1}(\mathbb{R})$ for $s < \frac 12$.
You can compute the double integral defining the norm and find that it belongs to $W^{s,p}$ for $s<1/p$. For $p=2$ you can also use the Fourier transform.
Up to some normalization, the Fourier transform of the characteristic function $\mathbf 1_I$ of a compact interval $I$ is
$$
\widehat{\mathbf 1_I}(\xi)=\frac{\sin \xi}{\xi}.
$$
Obviously the function $\mathbf 1_I$ is in $L^2(\mathbb R)$ but also in $W^{s, 2}(\mathbb R)$ for any $s<1/2$ since for $0\le s<1/2$, we have
$$
\int_{\mathbb R}\left\vert\frac{\sin \xi}{\xi}\right\vert^2 \vert \xi\vert^{2s}d\xi<+\infty.
$$
The index $1/2$ is sharp since
$
\int_{\vert \xi\vert\ge 1}\left\vert\frac{\sin \xi}{\xi}\right\vert^2 \vert \xi\vert d\xi=+\infty.
$
Is it possible to evaluate the $W^{s,2}$ norm though, or at least give a bound for it in terms of the measure of the interval or something related?
Yes. The norm is translation invariant so you may assume that $I=(-a,a)$. Now use dilations to relate the seminorn of any $u$ to that of $u_{\lambda} (x)=u(\lambda x)$.
|
2025-03-21T14:48:29.920527
| 2020-02-22T09:09:13 |
353298
|
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"M. Winter",
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|
Stack Exchange
|
Definition of convex hulls via maximal sets of interior-disjoint simplices
Let the simplex cover of a finite set $\mathcal{P}\subset \mathbb{E}^n,\ n\,\le\, k:=\operatorname{card}(\mathcal{P})\,\lt\infty$ of points in Euclidean $n$-space of which no $n+1$ are co-hyperplanar, denote a maximal interior-disjoint set of $n$-simplices with every corner an element of $\mathcal{P}$.
Questions:
letting $\mathrm{H}_\Sigma$ denote the set of simplex-sides, that are not shared by two simplices,
is $\mathrm{H}_\Sigma$ an invariant after a change of the simplex covers of a fixed pointset $\mathcal{P}$?
is $\mathrm{H}_\Sigma$ identical to the set of facets of the convex hull $\mathrm{CH}(\mathcal{P})$?
The reason for asking is the idea to base the definition of geometric hulls of discrete pointsets on the simplex-faces that are not shared by simplices by certain "regular" simplex-coverings of a pointset, provided the set of those undhared sides is invariant among all admissible coverings.
As there seem to be different definitions of the convex hull of a finite set of points in Euclidean spaces, in this context it shall mean what convex hull algorithms of computational geometry determine, namely the set of faces that constitute to the boundary of the intersection of all closed half-spaces in which $\mathcal{P}$ is contained.
Is $H_\Sigma$ really the set of simplex facets (a set of subsets of $\Bbb E^n$), or the union of these? In the former case, I think already the cube can be decomposed in several different ways into simplices. And just a minor note: in the latter case, you probably meant to ask whether $H_\Sigma$ is the boundary of the convex hull, as the union of the simplex facets is not convex (but the convex hull is).
@M.Winter thanks for the feedback; I have improved my question based on it. In one point I however disagree: the convex hull itself is the set of boundary points of the set of all closed half-spaces that contain $\mathcal{P}$ and, whereas that intersection is convex, it's boundary is not.
I see. It might be worthwhile to include your definition of convex hull into the question. I assumed the one from Wikipedia, that is, the convex hull as the smallest convex set containing these points.
@M.Winter thanks again for your valuable feedback!
|
2025-03-21T14:48:29.920720
| 2020-02-22T09:17:26 |
353299
|
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|
Stack Exchange
|
Internal vs. external definability of inner models
Suppose $\kappa$ is an inaccessible cardinal. Is the following situation consistent?
There is $p \in V_\kappa$ and a formula $\phi(x)$ such that there is exactly one $M \subseteq V_\kappa$ such that $M$ is a transitive set of size $\kappa$ and $M \models \phi(p)$.
The $M$ above is not a definable class in $V_\kappa$, meaning there is no $q \in V_\kappa$ and formula $\psi(x,y)$ such that $M = \{ x \in V_\kappa : V_\kappa \models \psi(x,q) \}$.
This seems to be easy to achieve given appropriate system of switches. Assume we have $V_\kappa$, a formula $\psi (\alpha)$ and transitive sets $\langle M_A\subseteq V_\kappa \mid A\subseteq \kappa\rangle$ such that for any $A\subseteq \kappa$ the mode $M_A$ is the unique transitive subset of $V_\kappa$, for which $M_A\models \varphi(\alpha)\iff \alpha\in A$, for $\alpha<\kappa$. Additionally assume that the satisfaction relation for $L_\kappa$ isn't definable in $V_\kappa$. Now the idea is to consider $M=M_A$, where $A$ encodes the satisfaction relation for $L_\kappa$.
Is $M$ a model of ZF or ZFC, or is it just a class satisfying this formula?
@AsafKaragila The example in mind was where $M$ is a ZFC model, but the more general question is interesting.
I'm asking because Johannes in his answer assumes the model satisfies some fragment of ZFC, which is a consequence of assuming ZFC. But that fragment seems to depend on the formula, so it's not too uniform.
I will try to partially answer your question. I claim that if $\kappa$ is weakly compact then this situation is inconsistent: I will show that such an $M$ is necessarily definable in $V_\kappa$.
Define the finite set of formulae $\Lambda:=\{\phi(x)\} \cup \text{tc}(\phi(x))$, where tc denotes the transitive closure (with respect to the 'proper sub-formula' relation). Use reflection in $M$ to find a transitive $q \prec_{\Lambda} M$ such that $p \in q$ and $q \in M$.
For every $x \in V_\kappa$ we shall inductively construct a ${<}\kappa$-branching tree $T_x$. It will consist of sequences (always with a last element) of transitive, $\Lambda$-elementary submodels containing $x$ and satisfying $\phi(p)$:
Set $\langle q \rangle$ to be the root of $T_x$. Assume that $\bar{y} \in T_x$ and $\bar{y}=\langle y_0, ... , y_\alpha\rangle$. Define the set of successors of $\bar{y}$ in $T_x$ as follows: $$\text{succ}_{T_x}(\bar{y}):=\{\bar{y}^{\frown} z \,\,\colon \,z \in V_{f(\bar{y})} \land y_\alpha \prec_{\Lambda} z \land x, y_\alpha \in z \land z \, \text{is transitive} \, \}$$ where $f(\bar{y}):=\max(\vert y_\alpha \vert^+ , \vert x\vert^+)$. In the limit case let $(\bar{y}_\alpha)_{\alpha < \gamma}$ ($\bar{y}_\alpha$ has length $\alpha +1$) be an increasing chain and set $\bar{y}_\gamma:=\langle y_0,...,y_\alpha,... \rangle ^\frown y_\gamma$, where $y_\gamma:= \bigcup_{\alpha < \gamma} y_\alpha$.
I claim that $x \in M \Longleftrightarrow T_x \,\, \text{has height} \,\, \kappa$.
Assume that $x \in M$. Using reflection in $M$, Löwenheim-Skolem and Mostowski collapse (and $q \in M$) one can easily show that $T_x$ must have height $\kappa$.
On the other hand, assume that $T_x$ has height $\kappa$. Any branch through $T_x$ defines an increasing chain of transitive, $\Lambda$-elementary submodels. The direct limit of this chain is a transitive model $M' \subseteq V_\kappa$ of size $\kappa$, satisfying $\phi(p)$ and containing $x$. By your assumption $M'=M$ follows and so $x \in M$.
Ok, I am also assuming that M satisfies enough of ZFC* so that reflection is actually possible inside M.
|
2025-03-21T14:48:29.920978
| 2020-02-12T13:21:45 |
353301
|
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|
Stack Exchange
|
Bound on the number of bridges between vertices in a sampled subgraph
I am researching connectivity in sampled subgraphs and have come across the following problem.
A bridge between two vertices $a$ and $b$ of a graph $G$ is an edge $e$ such that removing $e$ from $G$ disconnects $a$ and $b$. Note that $e$ is not a bridge if $a$ and $b$ are not connected in $G$. Denote by $B_G(a, b)$ the number of bridges between $a$ and $b$.
I am interested in how the number of bridges between $a$ and $b$ asymptotically distributes when sampling a random subgraph of $G$. More concretely, let $G=(V, E)$ be a graph on $n$ vertices, possibly with multiple edges. Now, sample each edge of $G$ independently with probability 1/2, yielding a new graph $G'$.
My question is: Is there a non-trivial function $g(n)$ and an $\ell>0$ such that
$$P(\forall a, b\in V(G)\colon B_{G'}(a, b)<g(n))>1-n^{-\ell}$$
for every choice of base graph $G$?
One example to note is the following. Sample the graph $G$ from $\mathcal G_{n, \frac 2n}$, i.e., the complete graph where each edge is sampled with probability $2/n$. Then the sampled subgraph $G'$ og $G$ is distributed according to $\mathcal G_{n, \frac1n}$. The latter has components of size $\Theta(n^{2/3})$ and I believe that each such component is fairly tree-like of diameter $\Theta(n^{1/3})$. In that case, this shows that we must have $g(n)= \Omega(n^{1/3})$.
This is true. However, I would like to have some bound that works for every graph. I have edited my question to include an example, where there will be many bridges.
|
2025-03-21T14:48:29.921099
| 2020-02-22T10:34:52 |
353302
|
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|
Stack Exchange
|
Construct $A_\infty$ bimodules maps from dg-maps
Let $ A $ be a dg-algebra. Let $U,V,W$ and $Z$ be dg-bimodules over $A$-$A$. Suppose I have cofibrant replacements $\pi_U : Up \rightarrow U$ (as right dg-module) and $\pi_W : pW \rightarrow W$ (as left dg-module). I can use $\pi_U$ to endow $Up$ with an $A_\infty$-bimodule structure over $A$-$A$, so that $\pi_U$ becomes an $A_\infty$ quasi-isomorphism, and similarly for $pW$. Also suppose $\pi_U \otimes 1 : Up \otimes_A V \xrightarrow{\simeq} U \otimes_A V$ is a quasi-isomorphism, and so is $1 \otimes \pi_W$. Thus I have $A_\infty$-bimodule maps $(\pi_U \otimes 1)^{-1}$ and $(1 \otimes \pi_V)^{-1}$ that are their quasi-inverses.
Suppose I have a map $f : Z \rightarrow Up \otimes_A V \otimes_A pW$ of complexes of graded vector spaces. If the composition $(\pi_U \otimes 1) \circ f : Z \rightarrow U \otimes_A V \otimes_A pW$ is a map of left dg-modules, then I can define a map $$f' := (\pi_U \otimes 1)^{-1} \circ ((\pi_U \otimes 1) \circ f) : Z \rightarrow Up \otimes_A V \otimes_A pW$$ of left $A_\infty$-modules, whose degree zero part is $f$.
Now suppose we also have that $(1\otimes \pi_W) \circ f' : Z \rightarrow Up \otimes V \otimes W$ is a map of right dg-modules. Then, I can define $$f'' := (1 \otimes \pi_W)^{-1} \circ ((1\otimes \pi_W) \circ f') : Z \rightarrow Up \otimes_A V \otimes_A pW.$$
If I understand this correctly, it should be both a map of left $A_\infty$-modules, and of right $A_\infty$-modules. However, this does not mean it is a map of $A_\infty$-bimodules, since I could need higher maps mixing both the left and the right action.
However, if I assume that $Z$ and $Up \otimes_A V \otimes W$ are concentrated only in homological degree 0 and 1, then all such higher maps must vanish. Thus, $f''$ should be a map of $A_\infty$-bimodules. Does that seem right ? And are there any other criterions that would allow me to conclude in the case where I don't have this degree hypothesis ?
More generally, I'm interested in finding ways to build maps of $A_\infty$-bimodules from $Z$ to $Up \otimes_A V \otimes pW$, from a map of graded complexes that gives maps of left or right dg-modules after composition with quasi-isomorphisms.
|
2025-03-21T14:48:29.921250
| 2020-02-22T10:46:21 |
353303
|
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|
Stack Exchange
|
Lie groupoids being homotopy equivalent
Let $M,N$be two smooth manifolds. Let $f,g:M\rightarrow N$ be two smooth maps. We have the notion of a homotopy (smooth homotopy) from the maps $f$ to the map $g$.
Is there a similar concept for morphisms of Lie groupoids?
Suppose $\mathcal{G}=(\mathcal{G}_1\rightrightarrows \mathcal{G}_0)$ and $\mathcal{H}=(\mathcal{H}_1\rightrightarrows \mathcal{H}_0)$ be Lie groupoids.
Let $(\phi_1,\phi_0),(\psi_1,\psi_0): \mathcal{G}\rightarrow \mathcal{H}$ be two morphisms of Lie groupoids. Is there any notion of a “homotopy” from $(\phi_1,\phi_0)$ to $(\psi_1,\psi_0)$?
Further, is there a notion of when two Lie groupoids are homotopy equivalent?
Is there a notion for topological groupoids?
Thank you @YCor
There might be more than one: do you want to consider that equivalences of groupoids are homotopy equivalences, or you want to keep the information of the space of objects ?
@SimonHenry I want to keep information on the space of objects.. I do not know if it is reasonable to call equivalence of Lie groupoids as Homotopy equivalence
Another source to learn about when two Lie groupoids are homotopy equivalent is the homotopy theory created by Joost Nuiten: https://link.springer.com/article/10.1007/s10485-019-09563-z. If I had more time, I'd try to dig into the paper to find the weak equivalences, but I'm swamped. So I just leave you the reference.
Yes there is! Here is one way to go.
If $X=(X_{1}\rightrightarrows X_{0})$ is a topological groupoid, then $X\times [0,1]=(X_{1}\times[0,1]\rightrightarrows X_{0}\times[0,1])$ is also a topological groupoid.
So the notion of homotopy is: if $f,f':X\rightarrow Y$ are two maps, then a homotopy between them is a map $F:X\times[0,1]\rightarrow Y$ that restricts to $f$ and $f'$ at $X\times\{0\}$ and $X\times\{1\}$.
And as soon as you have the notion of a homotopy of maps, you have the notion of a homotopy equivalence, defined as usual to be a pair of maps $f:X\rightarrow Y$ and $g:Y\rightarrow X$ and a pair of homotopies between the composites $fg$ and $gf$ and the respective identity maps.
I think I am missing something here...You are only saying about Homotopy equivalence of topological spaces... There is a +1 to your answer, so, there must be something interesting in your answer, it is just that I am misunderstanding
it might be confusing because it's formally identical to "homotopy equivalence" of topological spaces. But in the above all the objects are topological groupoids and all maps between them are maps of groupoids. Note that a topological space gives rise to a pretty canonical topological groupoid by declaring that there are no non-identity morphisms, i.e. the space $X$ becomes the groupoid $X\rightrightarrows X$.
Yes, that one I am aware of associating a topological groupoid for a topological space, Lie groupoid for a manifold.. Has this formal notion of Homotopy equivalence used any where else in literature, in similar way a notion of homotopy equivalence of topological spaces used?
Yeah, this is the standard notion of homotopy for groupoids. In fact this construction using the "cylinder object" $X\times[0,1]$ is the standard way to define "homotopy" in any model category see pg 233-234 of Quillen's "Rational Homotopy theory"
I believe that is the standard way to talk about Homotopy equivalences. But, I was thinking if we should consider the category of Lie groupoids whose morphisms are morphisms of Lie groupoids or that of bibundles (I have the set up of differentiable stacks in mind).. So, in that case should we involve the notion of bibundle (generalized morphism) when defining the notion of Homotopy equivalence?
@PraphullaKoushik you said in a comment on the question that you want to keep the space of objects. If you consider principal bibundles as morphisms instead of the morphisms in the answer, you only keep information about orbits of points and about their stabilisers.
@SebastianGoette So, can you please tell me what you feel could be another choice for the definition of Homotopy equivalence?
@PraphullaKoushik I am sorry, I am just learning groupoids myself. But here is my naive comment anyway: If you want to view a Lie groupoid with a single object as a Lie group $G$, then the version in the answer fits nicely. If you want to view the very same groupoid as a model for the classifying space $BG$ (as a differentiable stack), then you should use some version that is compatible with principal bibundles. It seems to me that both versions make sense, depending on the context.
@SebastianGoette Please let me know if you have any favorite reference for this set up?
|
2025-03-21T14:48:29.921556
| 2020-02-22T13:21:30 |
353307
|
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|
Stack Exchange
|
Is the formal completion of an affine group necessarily a formal group?
Notation and Setting: let $\operatorname{Aff}$ denote the category of affine schemes whose objects are covariant representable functors $\operatorname{X}:\operatorname{Ring}\rightarrow\operatorname{Set}$ and $\operatorname{Spec}:\operatorname{Ring^{op}}\rightarrow\operatorname{Func(Ring, Set)} $ be the contravariant Yoneda embedding of $\operatorname{Ring^{op}}$ in its category of presheaves so that $\operatorname{Aff}\simeq\operatorname{Ring^{op}}$.
In addition, let $\mathcal{O}:\operatorname{Func(Ring, Set)}\rightarrow\operatorname{Ring^{op}}$ be the functor that sends a functor $\operatorname{X}:\operatorname{Ring}\rightarrow\operatorname{Set}$ to the ring of maps $\operatorname{X}\rightarrow \mathbb{A}^1$ (where $\mathbb{A}^1$ is the forgetful functor) so that $\operatorname{Spec}$ and $\mathcal{O}$ are inverse of one another.
Let $\widehat{\operatorname{Aff}}$ be the indization of $\operatorname{Aff}$, i.e. the category whose objects are functors $\operatorname{X}:\operatorname{Ring}\rightarrow\operatorname{Set}$ that are small filtered colimits of affine schemes
Let $\operatorname{G}\in\operatorname{Aff}$ be a group object (affine group), i.e $\operatorname{(G(R),m_G)}$ is a group for every ring R, whereas $m_G$ is the multiplication map $m_G:\operatorname{G(R)}\times\operatorname{G(R)}\rightarrow\operatorname{G(R)}$ and $\mathcal{O}_G$ the ring of maps $\operatorname{G}\rightarrow\mathbb{A}^1$.
Let $I_0\supseteq I_1 \supseteq I_2 \supseteq \cdots$ be an infinite sequence of ideals in $\mathcal{O}_G$, i.e $I_n\in\mathcal{O}_G$ for every $n$. From Martin's argument we know that the formal colimit $\operatorname{Y}:=\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathcal{O}_{X}/I_n)}$ exists and is not an affine scheme (not representable).
My general question: When is $\operatorname{Y}$ a formal group, i.e. a group object in $\widehat{\operatorname{Aff}}$?
My considerations were as follows: since $\operatorname{Spec(\mathcal{O}_{G}/I_n)}\subseteq\operatorname{G}$ is a subfunctor for every $n$, the universal property of $\operatorname{Y}$ as a colimit gives us a unique map $\phi:\operatorname{Y}\rightarrow\operatorname{G}$ which implies that $\operatorname{Y}$ is a subfunctor of $\operatorname{G}$ (since every two parallel maps to Y which have the same image on G are equal to one another). In particular, $y\in\operatorname{Y(R)}$ corresponds to a map $f:\mathcal{O}_G\rightarrow R$ such that $I_n\subseteq\operatorname{Ker(f)}$ for some $n$. For every ring R, we thus have $\phi:\operatorname{Y(R)}\rightarrow\operatorname{G(R)}, y\mapsto y$.
This fact, and the example of $\operatorname{Nil}\simeq\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathbb{Z}[x]/(x)^{n})}$ which delivers $\operatorname{(Nil(R),+)}\subseteq\operatorname{(\mathbb{A}^1(R),+)}$ as a subgroup, lead me to the claim:
$\operatorname{Y}$ is a formal group if and only if $\operatorname{Y}$ is a sub group of $\operatorname{G}$
Now $\leftarrow$ is trivial. To show $\rightarrow$, we let $\operatorname{Y}$ be a formal group with multiplication map $m_Y$. Now if it is true (this is the point I am not sure about)
that $\phi$ is a map of formal groups and not just a normal natural transforamtion
then we have
$\phi(m_{Y}(y,y'))=m_{G}(\phi(y),\phi(y'))$ which, since $\phi(y)=y$, is equivalent to
$m_{Y}=m_{G}$ and we are done.
my specific questions:
Is the universal map $\phi$ necceserally a map of formal groups and if so why?
Assuming the answer to 1. is negative then:
2.1 In which cases is $\phi$ indeed a map of formal groups?
2.2 are there any examples of such situations where $\operatorname{Y}$ is not a subgroup of $\operatorname{G}$?
Edit: taking S. Carnahan's answer into consideration
Not assuming that $\phi:\operatorname{Y(R)}\rightarrow\operatorname{G(R)}, y\mapsto y$ is automatically a map of formal groups, it follows that that $\phi$ is a map of formal groups iff there is some $k$ such that the kernel of the map given by $m_G(y,y')$ lies in $I_k$.
Hopf ideals are kernals of maps of Hopf algebras. Thus
saying that $\operatorname{I}$ is a Hopf ideal is equivalent to saying that the map $\mathcal{O}_{G}\rightarrow\mathcal{O}_{G}/I$ is a map of Hopf algebras which is equivalent to the map $\operatorname{Spec(\mathcal{O}_{G}/I)}\rightarrow\operatorname{Spec(\mathcal{O}_{G})}$ being a map of affine groups.
My question: How do we show that the property $m_G(y,y')$ lies in a certain $I_k$ is equivalent to the fact that $I$ is a Hopf ideal?
Things to take into consideration:
We have that $I_k\subseteq I$.
We know (Milne Algebraic Groups Pro. 3.12., p.67) that, given a Hopf ideal $I$, any Hopf algebra homomorphism $A\rightarrow B$ whose kernal contains $I$ factors uniquely through $A\rightarrow A/I$
In the example of $\operatorname{Nil}$ we have for $a\in\operatorname{Nil_n(R)}, b\in\operatorname{Nil_m(R)}, (a+b)\in\operatorname{Nil_{n+m}(R)}$ and thus $k=n+m$. In addition, the map $\mathbb{Z}[x]\rightarrow\mathbb{Z}[x]/(x)\simeq\mathbb{Z}$ is a map of Hopf algebras whose kernal $(x)$ is hence a Hopf ideal. How to show the equivalence in this example?
Thanks
The universal map is not a map of formal groups without some extra condition. An easy class of counterexamples comes from completions of an affine group along a closed subscheme that does not contain the identity element. In general, when you want to complete an affine group to get a formal group, you set $I_n = I^n$ for an ideal $I$ that defines a closed subgroup. That is, $I$ is a Hopf ideal for the coordinate ring. In particular, $\Delta(I) \subseteq \mathcal{O}_G \otimes I + I \otimes \mathcal{O}_G$, so $\Delta(I_n) \subseteq \sum_{i+j=n} I_i \otimes I_j$.
To be more specific in the language you use, you need your system of ideals to be compatible with multiplication in the following way: If $y$ and $y'$ are $R$-points, such that the maps $\mathcal{O}_G \to R$ have kernel containing $I_n$ and $I_{n'}$, respectively, then there is some $k$ such that the kernel of the map given by $m_G(y,y')$ lies in $I_k$. You have not specified any structure that forces this condition to hold. For Hopf ideals, our calculation of $\Delta(I_n)$ shows that setting $k \geq n+n'$ is sufficient.
Here is an explicit example. We complete the additive group $\mathbb{G}_a = \operatorname{Spec} k[x]$ (for some nonzero commutative ring $k$) at the point $1$, so our ideals are $I_n = (x-1)^n$. Any $R$-point factors through some map $k[x]/(x-1)^n \to R$, but the product of such $R$-points comes from the composite $k[x] \to k[y,z] \to k[y,z]/((y-1)^m(z-1)^n) \to R$, where the first map takes $x$ to $y+z$, and hence $(x-2)^{m+n} \mapsto ((y-1)+(z-1))^{m+n}$. Then, this map necessarily factors through $k[x]/(x-2)^{m+n}$. Since $2 \neq 1$ in $k$, this map does not factor through $k[x]/(x-1)^r$ for any positive $r$.
Thanks a lot. I am almost there. I see how it all makes sense in the example of Nil and the Hopf ideal (x). What I still dont see is how you induce the inclusion of \Delta(I_n) \subseteq \sum_{i+j=n} I_i \otimes I_j from the defining inclusion of the Hopf ideal. I guess its simple tensor calculations but I am having some difficulties here.
@sagirot $\Delta(I^n) = \Delta(I)^n$ in a Hopf algebra.
@S.Carnahan This may be nit picky, but do you mean $ \mathbb{G}{m} $? I believe that $ \mathbb{G}{a} $ is the additive group, and so the identity is zero. The completion of $ \mathbb{G}{a} $ with respect to the point zero and the completion of $ \mathbb{G}{m} $ with respect to the point 1 are isomorphic over fields of characteristic zero via the exponential map, but it is probably best to use $ \mathbb{G}{m} $ for the linear alebraic group $ \operatorname{Spec}(k[t]{t}) $ which is isomorphic to $ (k,\times) $ as a group for clarity and consistency.
@S.Carnahan Everything else about your answer is on point.
@Schemer1 The condition missing from the question is that the completion must be at the identity. In my answer, I give an example of what happens when completing at the non-identity point $1 \in \mathbb{G}_a$. This is why I chose $\mathbb{G}_a$ instead of $\mathbb{G}_m$.
@S.Carnahan sorry, I missed that point and read hastily. In my hasty reading I thought that you were completing at the identity $ 1 \in \mathbb{G}{m} $ to form the formal group scheme $ \widehat{\mathbb{G}{m}} $. Thank you for clarifying.
|
2025-03-21T14:48:29.922051
| 2020-02-22T13:28:28 |
353308
|
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|
Stack Exchange
|
What is the number of connected graphs with $n$ vertices of max. degree up to $D$? Leaving $F(x) = x + x^2 + 2x^3 + 6x^4 + 21x^5 + 112x^6 + ...$
It is known that $F(x)$ is the generating function of the counting sequence of connected simple graphs with N vertices is given by:
$F(x) = x + x^2 + 2x^3 + 6x^4 + 21x^5 + 112x^6 + 853x^7...$
where the total number of connected graphs of $n$ vertices is obtained by the coefficient corresponding to $x^n$.
The question here is: How to find the series (or formula) to get the counting of connected simple graphs with $n$ vertices of maximum degree up to $D$? (generalizing with at least $1$ vertex with maximum degree equal to $D$)
For example, for $n = 5$ with $D=3$, we have $8$ connected graphs with at least $1$ vertex with maximum degree equal to $3$, and for the same $n$ with for $D=4$, we have $11$ connected graphs with at least $1$ vertex with maximum degree equal to $4$, as shown by the distribution of degrees in the figure below:
Follows a table of the count of the first groups of simple graphs according to the number of vertices and the maximum degree:
My question in brief is to determine the number of connected graphs with max. degree $D$ for a given number of Nodes $n$?
OBS: Do not get confused! the question does not include Disconnected Graphs, Digraphs, Labeled Graphs, number of cubic or quartic graphs!
Anyone know a solution to this intriguing problem? help me!
Note:
$F(x)$ is the generating function of the counting sequence of connected structures, then the corresponding generating function $G(x)$ of the counting sequence of all structures is given by:
$$G(x) = exp \sum_{k>=1} F(x^k)/k = \sum_{i=0} b_i x^k$$
Applying a variant of the Mobius inversion to equation above, it is possible to
express $F(x)$ in terms of $G(x)$:
$$F(x) = exp \sum_{k>=1} [\mu(x)/k]log G(x^k) = \sum_{i=0} a_i x^k$$
Where $\mu$ stands for the Mobius function.
Let $a_n$ be the number of connected graphs on $n$ vertices and $b_n$ the number of all graphs on N vertices. Then:
$F(x) = x + x^2 + 2x^3 + 6x^4 + 21x^5 + 112x^6 + 853x^7 + ...$
and
$G(x) = 1 + x + 2x^2 + 4x^3 + 11x^4 + 34x^5 + 156x^6 + ...$
I saw this on "Counting Disconnected Structures: Chemical Trees, Fullerenes, I-graphs" on the link: https://hrcak.srce.hr/file/4232 but I did not understand how I can use it in favor of my problem.
An observation:
If $T(n)$ represents the total of connected graphs of $n$ nodes and $D_p(n)$ represents the total of connected sub-graphs group with max. degree $P$, and knowing that $P_{max} = (n-1)$, we can write:
$$ T(n) = D_1(n) + D_2(n) + D_3(n) + ... + D_{n-1}(n)$$
For $n>2$, we know that $D_1(n) = 0$ and $D_2(n) = 2$, so we have to:
$$T(n) = 2 + D_3(n) + ... + D_{n-1}(n)$$
Note also that $D_{n-1}$ can be obtained by the coefficient of $x^{n-1}$ of $G(x)$!
But how to calculate $D_p(n)$?
OBS: Another way to attack the problem I posted on the link: About counting the number of graphs by the maximum degree $D$.
It seems worth noting that you are counting isomorphism classes of graphs, not graphs in total. This usually makes counting harder (think, e.g., counting partitions vs ball-and-bin problems).
There is unlikely to be a formula as such, although as noted by others it can be computed from the values for not necessarily connected graphs. (See A263293 in OEIS for values up to 10 nodes - these values were obtained by brute force enumeration of all graphs). Better methods exist, since it isn't necessary to enumerate every graph - only every degree sequence, but to my knowledge this has not been done. This technique, might give values up to about 15 nodes, but it is still exponential.
I agree with Andrew that a usable exact formula is unlikely except in special cases. An asymptotic formula for fixed $D$ would be not so difficult and perhaps it is published somewhere.
I think a general formula here does exist, but its complexity is similar to that of Jovovic's formula.
|
2025-03-21T14:48:29.922408
| 2020-02-22T13:49:13 |
353309
|
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|
Stack Exchange
|
Boundary map for Mayer-Vietoris sequence for Bordism
I am trying to reproduce some of the details in how the Mayer-Vietoris sequence for bordism should go, especially in showing exactness using this definition of the boundary operator. I have tried to work out some of the details, I'd be grateful to anyone who can take a look at them and perhaps fill in some of the gaps.
First recall that we want an exact sequence
$$\dots \rightarrow \Omega_{n+1}(X)\xrightarrow{\delta_{n+1}} \Omega_{n}(U\cap V)\rightarrow \Omega_{n}(U)\oplus \Omega_{n}(V)\rightarrow \Omega_{n}(X)\rightarrow \dots$$
Here's how I guess the $\delta_{n+1}$ should be defined. One takes a map $f:W\rightarrow X$ representing an element of $\Omega_{n+1}(X)$ and chooses a function $\phi:X\rightarrow [-1,1]$ such that $\phi|_{X\setminus U}=-1$ and $\phi_{X\setminus V}=1$. Then, using Thom transversality, one homotopes the composition $\phi\circ f$ to a smooth map $\tilde{\phi}$ that is tranverse onto the point 0. That way one can take the inverse image $M:=\tilde{\phi}^{-1}(0)$ and this will be a n-dimensional submanifold of $W$ with a map $f|_{M}:M\rightarrow U\cap V$. We get that it indeed maps to $U\cap V$ because we could choose the map $\tilde{\phi}$ to be arbitrarily close to $\phi\circ f$. The choice of $\delta_{n+1}$ is independant of the choice of $\tilde{\phi}$, as any two choices of $\tilde{\phi}$ will be homotopic and hence give bordant classes.
Now to show exactness at $\Omega_{n}(U\cap V)$ we want to show that the compositions $M\rightarrow U\cap V \rightarrow U$ and $M\rightarrow U\cap V\rightarrow V$ are null-bordant. For this we can simply take the submanifold $\tilde{\phi}^{-1}([0,1])$. This is then a manifold with boundary $\tilde{\phi}^{-1}(0)=M$ and hence this gives us our null-bordism of $M\rightarrow U$ and similarly for $M\rightarrow V$ we can take the null-bordism $\tilde{\phi}^{-1}([-1,0])$.
The main problem for me begins with the reverse direction of exactness. Starting with a representative $M\rightarrow U\cap V$ such that $M\rightarrow U\cap V\rightarrow U$ and $M\rightarrow U\cap V\rightarrow V$ are null-bordant, say by null-bordisms $W_1\rightarrow U$ and $W_2\rightarrow V$, we can paste together these manifolds along the common boundary $M$ to get a representative $[W_1\cup_{M} W_2\rightarrow X]\in \Omega_{n+1}(X)$. What I can't figure out is then how to show that this represents maps back to $M\rightarrow U\cap V$ under the map $\delta_{n+1}$.
Anyone have any ideas/references for this? I'd be very grateful for any help!
ok, coincidentally I have managed to find a reference after having randomly clicked on some link a few minutes after posting that made some reference to a book by Tom Dieck. And indeed there it is, on page 523, what I am looking for! (https://www.maths.ed.ac.uk/~v1ranick/papers/diecktop.pdf)
You have to be a bit more careful in your argument since your statements about transversality require smoothness or some other hypotheses (or fancier theorems). In general, this kind of issue can make it hard to show that certain bordism theories really are homology theories, at least by hands on arguments.
@GregFriedman, that part is guaranteed by the statement of the transversality theorem (such as on pg 24, appendix 2, Stong's "Notes on Cobordism theory"). It homotopes the continuous map $\phi\circ f$ to a smooth map $\tilde{\phi}$ that is moreover transverse regular on 0, so that the inverse image is indeed a submanifold of codimension 1. I have fixed this in the question, now
|
2025-03-21T14:48:29.922649
| 2020-02-22T14:16:56 |
353312
|
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|
Stack Exchange
|
Regularity on the boundary for the heat equation with linear source
This is probably a known problem but I was not able to find exactly what I am looking for.
I have the following linear heat equation with zero-flux boundary conditions:
\begin{equation}
\begin{cases}
\dot{u} - \Delta u = u \quad \text{in} \quad \Omega;\\
\nabla u \cdot \boldsymbol{n} = 0 \quad \text{on} \quad \partial \Omega,
\end{cases}
\end{equation}
on a NON-CONVEX polygonal domain $\Omega \subset \mathbb{R}^2$, with $\Delta$ being the Laplace operator. From basic energy arguments, we have the following estimate
\begin{equation}
\tag{1}
\label{true_estimate}
\|u(t)\|_{H^1(\Omega)} \leq C \|u(0)\|_{H^1(\Omega)} \exp(Ct), \qquad t >0.
\end{equation}
Question: if $\Gamma \subset \partial \Omega$ is an edge of $\Omega$ -or a portion of an edge to avoid corner singularities-, do we have a similar estimate for the trace of $u$ on $\Gamma$? That is:
\begin{equation}
\tag{2}
\label{desired_estimate}
\|Tr(u(t))\|_{H^1(\Gamma)} \leq C \left(\|Tr(u(0))\|_{H^1(\Gamma)} + \|u(0)\|_{H^1(\Omega)}\right)\exp(Ct), \qquad t>0.
\end{equation}
If \eqref{desired_estimate} does not hold true, could you please provide a counterexample.
Challenges
Estimate \eqref{desired_estimate} cannot be inferred from \eqref{true_estimate} via trace theorem because the trace of a $H^1(\Omega)$ function is only $H^{1/2}(\Gamma)$. So we try to exploit higher regularity in the interior.
But we don't have enough regularity in the interior due to non-convexity. The best we can expect is a $H^{1+\varepsilon}(\Omega)$ estimate, with $0< \varepsilon < 1/2$. So, by the trace theorem we could get a $H^{1/2 + \varepsilon}(\Gamma)$ estimate on the boundary, with $0< \varepsilon < 1/2$.
The estimate $(2)$ is false even in a half-plane. Indeed, let $w$ be any solution to the heat equation on $\mathbb{R}^2 \times [0,\,\infty)$ that is even in $y$ (so $w$ solves the Neumann problem for the heat equation in the upper half-plane), and vanishes on the $x$-axis at $t = 0$. Then $w_R(x,\,y,\,t) := w(Rx,\,Ry,\,R^2t)$ solves the Neumann problem for the heat equation in the upper half-plane, vanishes on the boundary at $t = 0$, and has initial $H^1$ norm independent of $R$ (by the two-dimensionality of the spatial domain). We have in addition that
$$f_R(t) := \int_{\mathbb{R}} |\partial_x w_R|^2(s,\,0,\,t)\,ds = Rf_1(R^2t).$$
Thus, if we let $u_R = e^tw_R$, then $u_R$ solves the desired Neumann problem in the upper half-plane, but at $t = R^{-2}$ the left side of $(2)$ is larger than $\sqrt{f_R(R^{-2})}\sim R^{1/2}$ and the right side is $\sim 1$ (independent of $R$). Taking $R \rightarrow \infty$ we see that such an estimate cannot hold.
Thank you very much. Your argument also implies that estimate (2) can fail regardless of the regularity class of the initial condition right? However, does (2) become true by accounting for the $H^2(\Omega)$ norm of the initial datum? For instance, in your specific example the $H^2$ norm of $w_R$ is not bounded uniformly in $R$. Is that correct?
That is right, the initial condition in the example can be smooth, and if you replace the $H^1(\Omega)$ norm with the $H^2(\Omega)$ norm on the right side of $(2)$ then the estimate follows from trace inequalities.
|
2025-03-21T14:48:29.922869
| 2020-02-22T15:38:29 |
353317
|
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|
Stack Exchange
|
Nonbraided rigid monoidal category where left and right duals coincide
In a braided rigid monoidal category $(\mathcal{M},\otimes)$ left and right duals coincide. What is an example of a rigid monoidal category where left and right duals coincide but there exist no braiding for the category?
Modules over an involutive Hopf algebra, which is not quasitriangular.
It may be useful to know that the standard term for when left and right duals coincide in a coherent way is "pivotal category". Any pivotal category which is braided is automatically spherical. So any non-spherical pivotal category will give an example which does not have any braiding.
@Bugs Bunny: Does left and right duals coinciding in $_H-mod$ imply the Hopf algebra $H$ s quasi-triangular?
@Fofi Konstantopoulou No way.
Adding to Tobias Fritz' comment, it's worth mentioning that a braided rigid category is not pivotal in general. Left and right duals a priori only coincide as functors, but not as monoidal functors, which is really the compatibility condition one would like.
The simplest example is G-graded vector spaces where G is a non-abelian group.
Is it clear that this category does not admit a braiding?
The fusion rules aren't commutative!
Yes, of course!
|
2025-03-21T14:48:29.922977
| 2020-02-22T18:03:26 |
353324
|
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|
Stack Exchange
|
Norm-controlled inverses vs uniform openness of multiplication
Let $A$ be a unital commutative Banach algebra and let $\hat{a}\in C(\Phi_A)$ be the Gelfand transform of an element $a\in A$. The algebra $A$ has norm-controlled inverses, whenever there exists a function $h\colon [0,\infty)^2\to [0,\infty)$ so that
$$\|a^{-1}\| \leqslant h(\|a\|, \|\widehat{a^{-1}}\|_\infty)$$
for any invertible element $a\in A$. Many naturally arising cBas have this property, however, $A=\ell_1(\mathbb Z)$ (with convolution) is a notable example of a Banach algebra that fails to have norm-controlled inverses (see Nikolski's paper). Hence my question:
I wonder if there is any relation between norm-controlled inverse and having uniformly open multiplication. It is known that $\ell_1(\mathbb Z)$ does not have uniformly open convolution, however, ${\rm BV}[0,1]$ does have norm-controlled inverses, yet the (pointwise) product is not uniformly open (see Kowalczyk–Turowska).
Suppose that $A$ is a unital semi-simple commutative Banach algebra that has uniformly open multiplication. Does it have norm-controlled inverses?
Some examples of (complex) Banach algebras that have uniformly open multiplication:
$C(X)$ for a compact, zero-dimensional Hausdorff space (Komisarski in the real case but the proof works also for the complex scalars) and $C[0,1]$ (Behrends, unpublished in the uniformly open case). I have a proof that extends this to all $C(X)$ for $X$ metric and having covering dimension 1. When $\dim X \geqslant 2$, $C(X)$ fails to have open multiplication.
$C^1[0,1]$ does have norm-controlled inverses, and it seems I can prove it has uniformly open multiplication too.
References:
A. Komisarski, A connection between multiplication in $C(X)$ and the dimension of $X$, Fund. Math. 189 (2) (2006), 149–154.
S. Kowalczyk, M. Turowska, Multiplication in the space of functions of bounded variation. J. Math. Anal. Appl., 472 (2019), 696–704.
N. Nikolski, In search of the invisible spectrum. Ann. Inst. Fourier (Grenoble), 49 (1999), 1925–1998.
Hi Tomek, could you remind us of some examples which are known to have uniformly open multiplication? My understanding, after a very quick look at your paper with Szymon, is that C(K) need not have this property, but does have this property if K is zero-dimensional and compact. What happens for uniform algebras, or algebras such as C^1[0,1]?
@YemonChoi, sure, added. As for other uniform algebras than $C^1[0,1]$ I am not sure. A necessary condition for openness is that the maximal ideal space is (at most) one-dimensional.
|
2025-03-21T14:48:29.923142
| 2020-02-22T20:28:29 |
353331
|
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|
Stack Exchange
|
On the role of $\diamondsuit$
The well-known axiom $\diamondsuit$ states that there is a sequence $\langle A_\alpha:\alpha<\omega_1\rangle$ (a $\diamondsuit$-sequence) of countable sets with the property that for any $A\subseteq\omega_1$, there is a $\delta<\omega_1$ (equivalently, stationarily many $\delta<\omega_1$) for which $A\cap\delta= A_\delta$.
$\diamondsuit$ holds in the constructible universe $L$, and implies the Continuum Hypothesis. The axiom is also a key ingredient in many constructions, and research over the past half-century has given us a good understanding of when the use of $\diamondsuit$ is necessary, in the sense that it cannot be replaced by simply assuming the Continuum Hypothesis.
There are many strengthenings of $\diamondsuit$ that have been studied and utilized in constructions, e.g. $\diamondsuit^*$, $\diamondsuit^+$, $\diamondsuit^\sharp$, and of course $V=L$).
My question is: what are some examples of statements $\Phi$ where a strengthening of $\diamondsuit$ is known to imply $\Phi$, but it is still unknown if $\diamondsuit$ itself implies $\Phi$? How fertile is this terrain?
Baumgartner proved that $\diamondsuit^+$ implies the existence of a club minimal Aronszajn tree. It is known that CH is not enough and it is open whether $\diamondsuit$ is sufficient.
Thanks! What's the reference?
Baumgartner's result appears in "Order types of real numbers and other uncountable orderings" (MR0661296) and cited also in Moore's paper MR2369944 in which the consistency of CH + no club-minimal Aronszajn trees is proved. Later in Soukup's paper MR4013972, Soukup shows that it is possible to get CH + there is a Suslin tree + no club-minimal Aronszajn trees, and states that the problem of obtaining the club-minimal Aronszajn tree just from diamond is still open.
There are various applications of strong diamond principles to the normal vs. collectionwise Hausdorff problem. For example, Shelah proved that under $\Diamond^$ every normal first-countable space is $\omega_1$-collectionwise Hausdorff. However, I don't know whether the $$ can be dropped. Paul Szeptycki might know.
|
2025-03-21T14:48:29.923314
| 2020-02-22T20:37:08 |
353333
|
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|
Stack Exchange
|
Subrings of Jacobson rings
Suppose $A\subset B$ is an inclusion of commutative rings with $B$ Jacobson.
If $B$ is finitely generated as an algebra over $A$ does it follow that $A$ is Jacobson?
If $B$ is finitely generated as a module over $A$ does it follow that $A$ is Jacobson?
For the module-finite (i.e. second) question:
Yes, it is true: The point is that $A \subseteq B$ is integral, and the claim is true for all integral extensions.
When $A \subseteq B$ is integral, every prime $\mathfrak{p} \subseteq A$ comes as $\mathfrak{P}\cap A$ for a prime $\mathfrak{P} \subseteq B$ ("lying over theorem"), and morever, if $\mathfrak{P}$ is maximal then so is $\mathfrak{p}$ (e.g. by "going up theorem").
So given a prime $\mathfrak{p}$ of $A$, choose such lift $\mathfrak{P} \subseteq B$. By $B$ being Jacobson one has $\mathfrak{P}=\bigcap_{\mathfrak{P}\subseteq{\mathfrak{M}}\subseteq_{\mathrm{max}}B}\mathfrak{M}$. Intersecting this with $A$ yields $$\mathfrak{p}=\bigcap_{\mathfrak{P}\subseteq{\mathfrak{M}}\subseteq_{\mathrm{max}}B}(\mathfrak{M}\cap A)$$
where all the ideals $\mathfrak{M}\cap A$ are maximal in $A$. Thus, $A$ is Jacobson.
Dear Pavel, maybe it is worth emphasizing from the beginning that your elegant argument proves the result for an arbitrary integral overring $B$ of $A$, not only for finite ones. At my first superficial reading of your post I thought that integrality was only used as a step in the proof, whereas it suffices for the whole proof to go through. Anyway, bravo! (And +1, of course.)
@GeorgesElencwajg Dear Georges, thank you, I have edited the answer as you proposed.
Well done Pavel: I hope your answer will get the many more upvotes it deserves
A counterexample for the first question is any DVR $R$. Clearly, $R$ is not Jacobson. But if $\pi$ is the uniformizer, then $Q(R) = R[\frac{1}{\pi}]$ is a finitely generated $R$-algebra and a field, hence Jacobson.
I would not be surprised if the second statement is true, though.
|
2025-03-21T14:48:29.923481
| 2020-02-22T21:20:36 |
353335
|
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|
Stack Exchange
|
Eigenvalues of tridiagonal symmetric matrix
Could you tell me please, are there any analytical methods how to find eigenvalues of matrix such this one?
$$
\begin{pmatrix}
a_1 & b_1 & 0 & 0 & 0 & \ldots & 0 \\
b_1 & a_2 & b_2 & 0 & 0 & \ldots & 0\\
0 & b_2 & a_3 & b_3 & 0 & \ldots & 0 \\
0&0 &b_3 & a_4&b_4 & \ldots & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & \ldots & a_{n-1} & b_{n-1}\\
0 & 0 & 0& 0& \ldots & b_{n-1} & a_n
\end{pmatrix}
$$
I've found previously solutions for some special cases, but here all matrix elements are different and nonzero.
Any symmetric matrix can be brought to tridiagonal form through finitely many explicit steps. So if there were any explicit analytical solutions for the eigenvalues of a tridiagonal matrix, they would also apply to ALL symmetric matrices. Thus I think you're asking for too much!
I am not sure what's the exact meaning of "analytic" in "analytic methods".
If you expand $|A-\lambda I|$ in the last row or column twice, you obtain a "three term recurrency" for the characteristic polynomials. Polynomials satisfying this type of recurrencies
have been studied VERY much, and they have many remarkable properties. They are called orthogonal polynomials.
The literature on these matrices and polynomials is really enormous. There are few cases which can be solved "explicitly".
See, for example, Gantmakher and Krein, Oscillation matrices and kernels and small vibrations of mechanical systems, AMS Chelsea Publishing, Providence, RI, 2002.
MR2743058 Simon, Barry Szegő's theorem and its descendants. Princeton University Press, Princeton, NJ, 2011.
N. I. Akhiezer, Classical moment problem, Hafner Publishing Co., New York 1965.
Yes, the case in question is probably too general. Thanks for the answer.
Here is an elementary resource which gives details of a method along with nice examples.
http://homepage.divms.uiowa.edu/~atkinson/m171.dir/sec_9-4.pdf
Thanks for the answer and for the resource.
|
2025-03-21T14:48:29.923983
| 2020-02-22T21:21:54 |
353336
|
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|
Stack Exchange
|
An upper bound for the G.C.D. of $\binom{a}{3}$ and $\binom{b}{3}$
I can't seem to find anything in the literature on how to estimate the g.c.d. of $\binom{a}{k}$ and $\binom{b}{k}$. In particular, I would like to know why $\gcd(\binom{a}{3}, \binom{b}{3})\leq b \binom{a-b}{3}$ for $a-b\geq 4$. I'm confident it's true (computer search).
Cross-posted at MSE here.
Here is a similar looking result.
Lemma. For any positive integers $a$ and $b$ satisfying $a-b\geq 2$,
$$\gcd\left(\binom{a}{3}, \binom{b}{3}\right)\qquad\text{divides}\qquad (a+b-2)\binom{a-b+1}{3}.$$
Proof. The statement follows readily from the identity
$$(2a-b-1)\binom{b}{3}-(2b-a-1)\binom{a}{3}\ =\ (a+b-2)\binom{a-b+1}{3}.$$
Very nice! (Not quite directly helpful but I upvoted it anyway).
I will hopefully prove this statement below modulo a finite check (I believe this will also work to show that $\gcd(\binom{b}{3}, \binom{a}{3}) \le \varepsilon b(a-b)^3$ for any $\varepsilon > 0$ except for a finite set of counterexamples).
Denote for simplicity $a - b = c$ and $\gcd(\binom{b}{3}, \binom{a}{3}) = d$.
First we will show that $d \ll c^5$. Indeed, each divisor of $d$ comes from one of the pairs $(a-k, b - l)$, where $k, l = 0,1, 2$. Each such pair gives us as a contribution a divisor of $a - b + l - k$ and there are five such numbers from $a - b - 2$ to $a - b + 2$. Note also that (up to a finite number of divisors like extra $6$ or so) different pairs with the same $l - k$ will contribute to the different prime divisors of $a - b + l - k $ because it will otherwise divide $\gcd(a - k, a - k_1) | (k - k_1)$. So we get $\ll (a - b - 2)\ldots (a - b + 2) \ll c^5$.
Therefore, if $c \le \delta \sqrt{a}$, then we are done. Indeed, in that case $d \ll c^5 \le c^3\delta^2 a$ while $b\binom{c}{3} \asymp ac^3$. Thus, $c \ge \delta \sqrt{a}$ for some fixed positive $\delta$.
Now assume that $c \ge C a^{2/3}$ for big enough $C$. Then $b\binom{c}{3} \gg C^3ba^2 \gg C^3b^3$ which is bigger than $\binom{b}{3}$ for big enough $C$ and we are done (here we used the obvious fact that $d\le \binom{b}{3}$). So $c \le Ca^{2/3}$ for some fixed positive $C$. In particular, $c = o(a)$ so $b \sim a$.
Now we will use the ingenious observation of GH from MO that
$$(2a - b - 1)\binom{b}{3} - (2b - a - 1)\binom{a}{3} = (a + b - 2)\binom{c + 1}{3}.$$
Note that if $\gcd(2a-b-1, 2b - a - 1) \ge C$ then we are done. Indeed, in that case we can divide by this $\gcd$ and get that $d \ll \frac{(a+b-2)c^3}{C}$. Since now $a\sim b$ we get that this is $\ll \frac{bc^3}{C}$ which is what we need for big enough $C$ (being a bit more careful we can show that it is enough to consider cases with $\gcd(2a-b-1, 2b-a-1) \le 2$).
We have $$\gcd(2a-b-1, 2b - a - 1) = \gcd(2a-b-1, 3(a-b)) \ge \gcd(2a-b-1, a-b)=\\ =\gcd(a-1, a - b) =\gcd(a-1, b-1).$$
Thus, $a-1$ and $b-1$ are almost coprime.
Similar to the above computation we can get that (up to the division by some uniformly bounded number) $d = (a+b-2)\binom{c+1}{3}$. We have here four factors: $(c-1), c, c+1, a+b-2$.
$c-1$ can come from gcd of $a-1,b$ or gcd of $a-2, b-1$.
$c$ can come from gcd of $a, b$; $a-2, b-2$ or $a-1, b-1$, but the last case can be excluded since $\gcd(a-1, b-1) \le C$.
$c + 1$ can come from gcd of $a, b-1$ or gcd of $a-1, b-2$.
Finally, $a+b-2$ can come from gcd of $a, b-2$; $a-2, b$ or $a-1, b-1$, but the last case we can again exclude.
Assume that $c-1$ splits between gcd of $a-1, b$ and gcd of $a-2, b-1$ as $r_1$ and $s_1$. Similarly for the next ones we would have $r_2, s_2$, $r_3, s_3$ and $r_4, s_4$.
Observe that different numbers among $r_1, \ldots , s_4$ corresponding to the same number among $a, a-1, a-2, b, b-1, b-2$ will be almost coprime in a sense that their $gcd$ is at most some constant (for $r_1, \ldots , s_3$ it is obvious because $c-k, c-l$ are almost coprime and with $r_4, s_4$ e.g. when we look at $r_1$ and $s_4$ corresponding to $b$ we have $\gcd(r_1, s_4) \mid \gcd(a-b-1, b, a+b-2) = 1$. Other cases are similar).
Note that $r_1s_1 \sim r_2s_2 \sim r_3s_3 \sim c$ and $r_4s_4 \sim a$.
Let's look at the numbers $b, b-2, a, a-2$:
For $b$ we have by almost coprimeness $r_1r_2s_4 \ll a$,
For $b-2$ we have $s_2s_3r_4 \ll a$,
For $a$ we have $r_2r_3r_4 \ll a$,
For $a-2$ we have $s_1s_2s_4 \ll a$.
Multiplying these inequalities we get $c^4a^2 \ll a^4$, that is $c \ll \sqrt{a}$.
So, after all this reasoning, we got that $\sqrt{a} \ll c \ll \sqrt{a}$ so $c\sim \sqrt{a}$! I guess one can reach the same conclusion from the assumption $d \ll \varepsilon b\binom{c}{3}$.
Now, we have two numbers which are (almost) divisible by our gcd: $\binom{c+2}{5}$ and $(a+b-2)\binom{c+1}{3}$. Note that both of them are proportional to the required $b\binom{c}{3}$. Thus, they are almost the same in a sense that $n\binom{c+2}{5} = m(a+b-2)\binom{c+1}{3}$ for some integers $n, m \le C$. Dividing by $\binom{c+1}{3}$ we get
$r(c+2)(c-2) = (a+b-2)$ for some $r\in \mathbb{Q}$, $r > 0$ with bounded denominator and numerator. So $a$ and $b$ are some second degree polynomials of $c$:
$$a = \frac{r(c+2)(c-2) + c + 2}{2}, b = \frac{r(c+2)(c-2)-c+2}{2}.$$
Therefore $\binom{a}{3}$ and $\binom{b}{3}$ are polynomials of degree $6$ in $c$. It remains to show that their $\gcd$ (as polynomials in $\mathbb{Q}[x]$) has degree at most $4$ – in that case $\gcd(\binom{a}{3}, \binom{b}{3}) \ll c^4$ while $b\binom{c}{3} \sim c^5$.
Since $\binom{a}{3}$ is a product of three quadratic polynomials, if $\gcd$ has degree at least $5$ then all factors should be used, in particular $a-1 = \frac{r(c+2)(c-2) + c}{2}$. If it is not coprime to some $b-k$ then their $\gcd$ divides their difference which is $\frac{2c + 2k - 2}{2}$. Thus, $c = 1 - k$ is a root of $a-1$ for $k = 0$, $k = 1$ or $k = 2$. Case $c = 0$ gives us $r = 0$, $c = -1$ gives us $r = -\frac{1}{3}$ and $c = 1$ gives us $r = \frac{1}{3}$. In all three cases the degree of gcd is at most $4$ and the proof is complete.
|
2025-03-21T14:48:29.924360
| 2020-02-22T21:43:28 |
353340
|
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|
Stack Exchange
|
What is a "cusp" ("кусок") in relation to Guba's embedding theorem?
I'm confused by the definition of a "cusp" as found in
V.S. Guba, Conditions for the embeddability of semigroups into groups, Math. Notes 56 (1994), Nos. 1-2, 763-769 (link).
In the words of Mark Sapir (from an answer that has meanwhile been removed from this thread), "cusp" is a «weird translation» into English of the Russian term "кусок" (piece) used by Guba in the original version of his paper. However, this is not the (main) point here: The point is rather that I have seen the same definition repeated elsewhere (with the terms "$s$-piece" or "$S$-piece" used in place of Guba's "кусок"), but continue to find it "strange". Let me try and explain my problem with Guba's definition, before asking a question.
Suppose that $S = \langle X \mid R \rangle$ is a semigroup presentation, and denote by $\mathscr F(X)$ the free monoid over $X$ and by $\ast$ the operation of word concatenation in $\mathscr F(X)$. Following Guba, it seems that
a defining word (relative to $S$) is, as usual, any word in the set $\bigcup_{(\mathfrak a, \mathfrak b) \in R} \{\mathfrak a, \mathfrak b\}$;
a cusp (кусок) is a word $\mathfrak u \in \mathscr F(X)$ such that there exist
$\mathfrak p, \mathfrak p^\prime, \mathfrak q, \mathfrak q^\prime \in \mathscr F(X)$, with $\mathfrak p \ne \mathfrak q$ or
$\mathfrak p^\prime \ne \mathfrak q^\prime$, such that $
\mathfrak p \ast \mathfrak u \ast \mathfrak q$ and $\mathfrak p^\prime \ast \mathfrak u \ast \mathfrak q^\prime$ are defining words.
But this doesn't look very natural. For one thing, what would prevent us from taking $\mathfrak p = \mathfrak p^\prime$ and $\mathfrak q = \mathfrak q^\prime$? Maybe I'm misunderstanding something and what Guba really means is that
a cusp (кусок) is a word $\mathfrak u \in \mathscr F(X)$ such that there exist
$\mathfrak p, \mathfrak p^\prime, \mathfrak q, \mathfrak q^\prime \in \mathscr F(X)$, with $\mathfrak p \ne \mathfrak q$ or
$\mathfrak p^\prime \ne \mathfrak q^\prime$, such $
(\mathfrak p \ast \mathfrak u \ast \mathfrak q, \mathfrak p^\prime \ast \mathfrak u \ast \mathfrak q^\prime) \in R$;
or maybe there is a typo in the paper. In fact, I checked the original version of Guba's article, and at least for what concerns the definition of a "cusp" ("кусок"), there hasn't been any typo introduced in the translation process. I've also tried to follow the proof of Theorem 1 in Guba's paper, but there are some details I'm still trying to demystify. On the other hand, it appears that Guba's notion of "cusp" ("кусок") is borrowed from E.V. Kashintsev's paper
Small cancellation conditions and embeddability of semigroups in groups, Int. J. Alg. Comp. 2 (1992), No. 4, 433-441;
except that Kashintsev uses the term "$s$-piece" instead of "cusp" ("кусок") and lets an $s$-piece be a word $\mathfrak u \in \mathscr F(X)$ for which there exist
$\mathfrak p, \mathfrak p^\prime, \mathfrak q, \mathfrak q^\prime \in \mathscr F(X)$, with $\mathfrak p \ne \mathfrak p^\prime$ or
$\mathfrak q \ne \mathfrak q^\prime$, such that $
\mathfrak p \ast \mathfrak u \ast \mathfrak q$ and $\mathfrak p^\prime \ast \mathfrak u \ast \mathfrak q^\prime$ are defining words (as remarked by Kashintsev, these two words need not be distinct). This is one reason making me think that there is a typo in Guba's definition. The following excerpt if from the English translation of Guba's paper:
The classes $K_p^q$ were studied in [1]. By definition, a semigroup $S$ belongs to the class $K_p^q$ if it can be specified by a corepresentation (1) that satisfies the small cancellation conditions $C_s(p)$ and $D(q)$.
Here, item [1] in the bibliography of Guba's paper is Kashintsev's paper (the same one mentioned in the above) and a semigroup is of class $K_p^q$ if it is isomorphic to a semigroup presentation $S$ with finitely many generators such that (i) no defining word can be expressed as the concatenation of less than $p$ "cusps" and (ii) the left and the right graphs of $S$ (that is, some undirected multigraphs naturally associated with the presentation) have both girth $\ge q$. Then, Guba's embedding theorem (that is, Theorem 1 in Guba's paper) is the statement that every semigroup of class $K_3^2$ embeds into a group. So I would expect Guba's notion of "cusp" ("кусок") to coincide with Kashintsev's notion of "$s$-piece", and yet the two definitions are different (and probably non-equivalent). With all this in mind, I'd like to ask the following:
Question. Can anyone familiar with these things clarify the situation or suggest a reference where Guba's embedding theorem or generalizations of it are discussed as part of a more systematic treatment of the subject (e.g., a chapter in a book concerned with the embedding of a semigroup into a group)?
Not sure what you mean by "For one thing, what would prevent us from taking p=p′ and q=q′?". The "piece" definition requires existence of a common subword "u" in 2 different defining words (or in the same def. word but then "u" should be contained at least twice in that word). This is what is written in the definition mentioned.
And yes, "cusp" is a very weird translation. A 'piece' is a standart term, it was used by E.V. Kashintsev in semigroup context in his paper in 1970. It was used even earlier for group representations.
@TT_ It's been a while, but I think it should be $\mathfrak p \ne \mathfrak p'$ or $\mathfrak q \ne \mathfrak q'$ rather than $\mathfrak p \ne \mathfrak q$ or $\mathfrak p' \ne \mathfrak q'$ (as explained in the OP).
@TT_ It seems to me that, by your interpretation of the text (nowhere in Guba's paper one can explicitly read that $u$ should be a common subword of two different def (= defining) words or contained twice in the same def word), every cusp (as per Guba's paper) is an $s$-piece (as per Kashintsev's paper), but not the other way around: Let $X$ be the $3$-element set ${a, b, c}$ and assume the only def rel in $R$ is $(a \ast b \ast a, c \ast b \ast c)$. Then $b$ is an $s$-piece (with $p=q=a$ and $p'=q'=c$); but is not a cusp because $b$ appears only in def words of the form $z \ast b \ast z$.
I would not call that my interpretation, that's rather standart description what the "piece" is (sorry, I can't make myself call it cusp). And yes, you are right - there is a typo in Guba paper. Being accustomed to the "piece" definition, it's hard to notice it.
Accordingly, both Guba and Kashintsev definitions are equivalent.
Also I think it's S-piece (related to presentation S) in Kashintsev paper. I don't have a copy of Kashintsev paper, but pretty sure he meant it as S-piece. I believe he used the word "piece" (in Russian of course) when I talked to him. Maybe in the paper he had to distinguish pieces of two different presentations, so used a prefix "S" for that?
@TT_ (i) I'm sticking to the term "cusp" only for the sake of discussion (you and others have fully convinced me that "cusp" was not a best choice in the translation from Russian). (ii) I have a copy of Kashintsev's paper Small cancellation conditions and embeddability of semigroups in groups and "$s$-piece" (with a lowercase "s") is the term being used therein. (iii) I should have bracketed the word "interpretation" in scare quotes: I hope it's clearer from the latest version of my own answer below what I was really meaning.
Update: I had an email exchange with Victor Guba. He has kindly confirmed that there is indeed a typo in (the Russian and English versions of) his paper: a "кусок" (as per his paper) and an "$s$-piece" (as per Kashintsev's paper) are meant to be one and the same thing.
The part below was written before hearing from Guba.
(A recent comment of user TT_ has reminded me of this question, which has also entered a private exchange with Laura Cossu these days. So I've thought to post an answer in the hope that it can also be useful to someone else or those more familiar than me with the subject can shed further light on the story.)
As far as I can tell, there are two possible interpretations of the facts described in the OP:
It is implicitly understood in Guba's paper (as suggested by TT_ in their comment under the OP) that either the defining words $w = \mathfrak p \ast \mathfrak u \ast \mathfrak q$ and $w' = \mathfrak p' \ast \mathfrak u \ast \mathfrak q'$ are distinct, or $\mathfrak u$ occurs twice as a subword of $w$.
There is a typo in Guba's paper (and the typo has unfortunately spread through the literature).
In the first case, a cusp (as per Guba's paper) is always an $s$-piece (as per Kashintsev's paper); and in both cases, an $s$-piece need not be a cusp. E.g., consider the monoid presentation $H := \langle X \mid R \rangle$, where $X$ is the $3$-element set $\{a, b, c\}$ and the only defining relation in $R$ is the pair $(a \ast b \ast a, c \ast b \ast c)$: The only $s$-pieces of $H$ are the empty $X$-word $\varepsilon_X$ and the generators $a$, $b$, and $c$; while, in any case, the only cusps are $\varepsilon_X$, $a$, and $c$ (in particular, $b$ is not a cusp because it only appears in defining words of the form $\mathfrak z \ast b \ast \mathfrak z$).
As I mentioned in a comment above, both interpretations are correct.
Also I have to mention that piece is usually defined as non-empty subword, there is no much sense to consider empty pieces. It should be mentioned in the definition of course.
Just for the sake of completeness, I'm not sure if it was Martin Greendlinger who first used the word piece. Maybe not. For ref.: https://en.wikipedia.org/wiki/Small_cancellation_theory
Sorry for so many comments... If needed, I can check/translate for you the definition from Kashintsev paper from 1970 where the diagram method was first applied to semigroups. His advisor was M.Greendlinger, so it was natural for them to extend the method to semigroup presentations.
@TT_ Many thanks for your informative comments. I agree that there is no much sense to consider empty pieces, but this possibility is not (explicitly) ruled out by the definitions in the papers cited in the OP and that's the only reason why I haven't ruled it out either.
|
2025-03-21T14:48:29.925078
| 2020-02-22T22:17:39 |
353342
|
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|
Stack Exchange
|
Inner automorphisms of group algebras vs. inner automorphisms of the group
In a recent question on MSE I asked about conditions under which the canonical morphism $Out(G) \to Out(k[G])$ is injective.
Is it true that this morphism is indeed injective if $G$ is finite and $k=\mathbb{Z}$ ?
What I already know:
It's not injective if $k$ is a field in non-modular characteristic. The kernel is $Out_c(G)$, the group of conjugacy class preserving automorphisms. That is Captain Lama's answer to my question.
It is true if $k=\mathbb{Z}$ and $G$ is nilpotent. That's my own answer to the question.
In the mean time I have also found out that the morphism is injective if $G$ is rational, i.e. if $g$ is cojugated to $g^k$ for every $g\in G$ and every $k\in\mathbb{Z}$ with $gcd(k,ord(g))=1$, e.g. $G$ a symmetric group.
Proof: If $\alpha\in Aut(G)$ is in the kernel of the morphism, say $\alpha(g)=ugu^{-1}$ for some $u\in\mathbb{Z}G^\times$, then
$$u^\ast u g=u^\ast \alpha(g) u = u^\ast \alpha(g^{-1})^{-1} u = (u^\ast \alpha(g^{-1}) u)^\ast = (u^\ast u g^{-1})^\ast = g u^\ast u$$
where $^\ast$ is the antiautomorphism of the group algebra with $g\mapsto g^{-1}$. Hence $u^\ast u \in\mathbb{Z}G$. We want to prove that $u^\ast u = 1$.
Consider all the complex characters $\chi\in Irr(G)$ and their central characters $\omega_\chi: Z(\mathbb{C}G) \to \mathbb{C}$. We can pick a matrix representations $\rho_\chi: G\to GL_n(\mathbb{C})$ affording $\chi$ with $\rho(G)\subseteq U_n(\mathbb{C})$ so that $u^\ast u$ is mapped to some self-adjoint and positive definite matrix so that $\omega_\chi(u^\ast u)\in\mathbb{R}_{>0}$. Furthermore $u^\ast u\in Z(\mathbb{Z}G)$ is integral over $\mathbb{Z}$ so that $\omega_\chi(u^\ast u)$ is an algebraic integer. Moreover it must be in $\omega_\chi(Z(\mathbb{Q}G)) =\mathbb{Q}(\chi)$. Now if $G$ is rational, then all characters have rational values so that $\omega_\chi(u^\ast u)$ is a real, positive, rational, invertible integer. In other words $\omega_\chi(u^\ast u) = 1$, i.e $\rho_\chi(u^\ast u) = 1_{n\times n}$. Since $\chi$ was arbitrary, $u^\ast u=1$.
The only units of $\mathbb{Z}G$ with $u^\ast u=1$ are elements of the form $\pm g$ so that $\alpha\in Inn(G)$ as we wanted. QED.
Note that we can get the same conclusion under weaker conditions. For example if $\mathbb{Q}(\chi)=\mathbb{Q}(i)$, then the only real, positive, integral unit is also 1.
Would you recall how Out of a $k$-algebra is defined?
Just as you'd think: $Out(A) := Aut_{k\text{-Alg}}(A) / Inn(A)$ where $Inn(A)$ is the normal subgroup consisting of automorphisms that are of the form $a\mapsto uau^{-1}$ for some unit $u\in A^\times$.
The question for finite $G$ and $k = \mathbb{Z}$ is the normalizer problem, see [1, Section 1]. By a result of Jan Krempa, the kernel of the cannonical morphism is in that case always an elementary abelian $2$-group. As far as I know, there is basically only one example known where the kernel is non-trivial [1, Theorem A]. This example was constructed by Martin Hertweck and used to provide a counterexample to the isomorphism problem for integral group rings.
[1] Martin Hertweck; A counterexample to the isomorphism problem for integral group rings. Ann. of Math. (2) 154 (2001), no. 1, 115–138, MR1847590.
Thank you. That's very interesting.
|
2025-03-21T14:48:29.925310
| 2020-02-22T23:36:28 |
353348
|
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|
Stack Exchange
|
Introductory text on amenability
I am looking for a book that covers amenability rigorously.
Preferably a book aimed at beginners.
Depending on what you want, Davidson's C$^$-algebras by example* or Wagon's The Banach-Tarski Paradox.
Appendix G on amenability of the book Kazhdan's Property (T) by Bekka, de la Harpe, Valette.
I suggest the book Amenability by Alan.t. Paterson. It contains everything about amenability of groups till the publication year 1988. But a newest one is Lectures on Amenability by Volker. Runde . It's main theme is about amenability of Banch Algebras.
|
2025-03-21T14:48:29.925394
| 2020-02-23T01:21:32 |
353351
|
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|
Stack Exchange
|
If $\|S\|<\sin\frac{\pi}{2n}$ then $\|P(I-S)^ku\|\neq 0$ for all $k=0,\ldots,n$
I want to show the following:
Let $H$ be a Hilbert space and let $S:H\to H$ be a bounded operator such that
$$\|S\|<\sin\frac{\pi}{2n}.$$
Let $\mathcal{L}$ be a closed subspace of $H$ and $$u_k:=(I-S)^ku,\;\;\;\;\text{for}\;\;\;k=0,\ldots,n\;\;\;\text{and}\;\;\;u\in\mathcal{L}\setminus\{0\}.$$
Prove that
$$\|Pu_k\|\neq 0\;\;\;\text{for}\;\;\;k=0,\ldots,n,$$
where $P$ denotes the orthogonal projection on $\mathcal{L}$.
An idea:
For $k=0$ is clear because $Pu_0=Pu=u\in\mathcal{L}\setminus\{0\}$.
For $k=1$, suppose that $Pu_1=0$ then $u_1\in\mathcal{L}^\perp$ and
\begin{align*}
0&=\langle u_0,u_1\rangle=\langle u_0,u_0\rangle-\langle u_0,Su_0\rangle\\
&\geq \|u_0\|^2-\|S\|\|u_0\|^2\\
&>\left(1-\sin\frac{\pi}{2n}\right)\|u_0\|^2>0
\end{align*}
which is not posible. So, $\|Pu_1\|\neq 0$.
For $k>2$, I don't know how to continue. Can someone give me an idea? Thanks.
What is the motivation for this question?
I am reading the article "Geometry of higher order realtive spectrum of linear operator in Hilbert spaces" by Eugene Shargorodsky. In Theorem 5.2 he shows that if $T$ is a bounded operator then $$\left(\sin\left(\frac{π}{2n}\right)\right)^{-1}|T|$$
is a bound of the $n$-th order spectrum $\mathrm{Spec}_n(T,\mathcal{L})$ for any closed linear subespace $\mathcal{L}$ of $H$. In his proof, he divides by the quantity $|Pu_k|$ but I do not know why this quantity is $\neq 0$.
Your condition that $\| S\|<\sin\frac{\pi}{2n}$ implies that the angle between
$u_{{k+1}}$ and $u_{k}$ is less than $\pi/2n$, for $k=0,...,n-1$. Therefore the angle
between $u=u_0$ and $u_{k}$ is $<\pi/2$ for $k=1,...,n$. Since $u\in L$, $u_1,...,u_n$ cannot be orthogonal to $L$ that is $Pu_k\neq 0$.
Nice solution, I wonder whether there is a proof without using properties of angles. By the way, the triangle inequality for angles is not straighforward and I found the original paper by Rao in the web. Any other source?
Triangle property for angles is easy. Since we are dealing with finitely many vectors, it is enough to consider their span, which is a finite dimensional space. Now the angle between vectors is the spherical distance on the unit sphere between the endpoints of the corresponding unit vectors.
You are right, I see. Then I wonder why there are papers on the subject!
Thank you for the reply. I do not underestand why the angle between $u_k$ and $u_{k+1}$ has to be less than $\frac{\pi}{2n}$. Also, if $H$ is a complex Hilbert space. How do I underestand the angle between two vectors in this span?
The angle is $\arccos$ of (the absolute value of the dot product, divided by the product of the norms).
With this definition, for the angle $\theta_k$ between $u_k$ and $u_{k+1}$ I did the following estimate:
$$\theta_k=\arccos\left(\frac{|\langle u_k,u_{k+1}\rangle|}{|u_k||u_{k+1}|}\right)\leq \arccos\left(\frac{1−\sin\frac{π}{2n}}{1+\sin\frac{π}{2n}}\right)$$
but
$$\arccos\left(\frac{1−\sin\frac{π}{2n}}{1+\sin\frac{π}{2n}}\right)>\frac{π}{2n}.$$ Is this correct?
Another question: Is the angle between $u_0$ and $u_k$ less than $\pi$ or less than $\pi/2$?. Thank you for your reply.
|
2025-03-21T14:48:29.925657
| 2020-02-23T02:03:31 |
353354
|
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}
|
Stack Exchange
|
Bounding an expectation involving i.i.d. standard Gaussians and Rademacher
I have tried to bound the following quantity, but cannot get the "right" (conjectured) bound:
$$
\phi(\gamma,d,n) = -1+e^{\frac{1}{2}n\gamma^2 d}
\mathbb{E}_{X}\left[\frac{\mathbb{E}_Z[\prod_{j=1}^n(1+\gamma\langle X^{(j)},Z\rangle)]^2}{\prod_{i=1}^d \cosh(\gamma\sum _{j=1}^nX^{(j)}_i)}\right]\tag{1}$$
where $\gamma\in(0,1]$, $d\gg 1$, $X=(X^{(j)})_{1\leq j\leq n}$ is a collection of i.i.d. standard $d$-dimensional Gaussian r.v.'s, and $Z$ is uniform on $\{-1,1\}^d$ (Rademacher) independent of the $X^{(j)}$'s.
Conjecture. Assuming $\gamma^2 d \leq 1$, as long as $n\gamma^4 d \ll 1$ we have
$$\phi(\gamma,d,n)\ll 1\tag{2}$$
I haven't been able to show it, even for $n=2$; there seems to be a delicate balancing act for things to exactly cancel the $e^{\frac{1}{2}n\gamma^2 d}$ factor...
Update: For (1) in the case $n=2$, my numerical experiments (a bit noisier than I had hoped) seem to be consistent with the conjecture:
$\gamma$ fixed, varying $d$
$d=5$ fixed, varying $\gamma$ (fit made with SciPy, in Python)
Update 2: For $d=1$ and $n\in\{1,2,3\}$, Mathematica could compute explicitly the expectation (though already it takes some time for $n=3$. The behavior on these few points is clearly linear wrt $n$:
Is $Y$ in $E_{XY}$ a misprint or some other r.v. $Y$ is, indeed, present?
@fedja Oh, good catch, that was a remain of what I wrote in an earlier version for $n=2$ (I had $X,Y$ instead of $X_1,X_2$). Edited.
One more stupid question: when $n=d=1$ we have $E_X(\frac {E_Z\dots}{\dots})=E_X(\frac {1+\gamma^2X^2}{\dots})\ge \frac{[E_X\sqrt{1+\gamma^2X^2}]^2}{E_X\dots}=e^{-\gamma^2/2}[E_X\sqrt{1+\gamma^2X^2}]^2$, so the whole expression is at least $[E_X\sqrt{1+\gamma^2X^2}]^2-1\sim \gamma^2$. How does this agree with your conjectured $\gamma^4$ decay as $\gamma\to 0$? Am I missing or misunderstanding something?
@fedja (from my phone) I am not sure how you get this denominator for $n=1$. Since the square is outside of the $\mathbb{E}_Z$, the denominator should be 1 (as $\langle X,Z\rangle$ has expectation zero), shouldn't it?
You mean "the numerator"? I see. $E_Z[\rm{big\ formula}]^2$ is ambiguous: I interpreted it as the expectation of the square rather than the square of the expectation. So it should be the square of the expectation in $Z$, right?
Yes, indeed (also, you're right, my above comment should read "numerator"). It's the square of the expectation (wrt $Z$). @fedja
I can also explicitly work out the cases $n=1$ and $(n,d)=(2,1)$ to get the $\gamma^4$ dependence, unless I made a mistake in my computations.
To the downvoter: is there something wrong with my question? How can I improve it?
For $d=1$ everything is trivially fine. The regime that bothers me is when both $d$ and $n$ are $c\gamma^{-2}$ with fixed (albeit small) $c>0$ and $\gamma\to 0$.
@fedja I agree $d=1$ is not the most... interesting regime. I just don't know how to approach the problem at the moment, and simulations with $d>1$ and $n > 1$ become computationally difficult quite fast.
@fedja The relation between $d$ and $\gamma$ you mention is the one I'm interested in, . Why do you think this regime (especially the one for $n$) is the possibly tricky one? (Note that for the "Gaussian analogue" of the question, where $Z$ is Gaussian and a few other details change accordingly, the $-1$ become something else which blows up to infinity when $n \nearrow \gamma^{-2}$)
Something looks fishy at least with the first conjecture. Perhaps I'm again misinterpreting something, but the argument is as follows:
Consider $n=2$. Write $X^{(1)}=\frac{X+Y}{\sqrt 2}, X^{(2)}=\frac{X-Y}{\sqrt 2}$. Then $X,Y$ are independent standard Gaussians in $\mathbb R^d$. Also
$$
E_Z[(1+\gamma\langle X^{(1)},Z\rangle)(1+\gamma\langle X^{(2)},Z\rangle)]=1+\gamma^2\langle X^{(1)},X^{(2)}\rangle
\\
=1+\frac{\gamma^2}2(\|X\|^2-\|Y\|^2)
$$
Now, since $\|Y\|^2$ is a sum of squares of $d$ independent Gaussians, it deviates from any fixed number by about $\sqrt d$ with constant probability, so the expectation $E_Y[E_Z[(1+\gamma\langle X^{(1)},Z\rangle)(1+\gamma\langle X^{(2)},Z\rangle)]^2]$ is at least $c\gamma^4d$ regardless of $X$ and the whole expression you are interested in is at least
$$
-1+c\gamma^4d\left(E[\cosh(\gamma \sqrt 2 W)]E[\frac 1{\cosh(\gamma \sqrt 2 W)}]\right)^d
$$
but for fixed $\gamma$ and $d\to+\infty$, this is, clearly, exponential in $d$. What am I missing this time?
I have to read carefully what you wrote (what is $W$ in the last line?),but I may have been making a hidden assumption there. Namely (as used in the last part of my answer, dealing with the Gaussian variant) I am implicitly assuming $\gamma^2 d$ is bounded (as I get $e^{\gamma^2 d}\cdot \gamma^4 d$ as an upper bound), which is clearly not an assumption stated in my original post, but is something I'll have in my end application. I'll check if, under that assumption, what you wrote still provides a counterexample.
Yes, under that extra assumption $\gamma2 d \lesssim 1$, the last part indeed stays bounded, so things are consistent. I apologize for wasting your time by not stating that upfront.
@ClementC. $W$ is the standard Gaussian. Sorry for not stating that. Yeah, if $\gamma^2d$ is bounded, then the exponential growth gets tamed. OK, let me try to work it out under that assumption but expect more stupid questions :-)
@ClementC. One comes right away: is $n$ bounded too? (if so, what's the point of the $n$ factor in the proposed equivalence; if not, then I'm afraid that the exponential factor may still prevail as $n\to\infty$)
n is not bounded itself, but making the assumption $n \gamma^4 d \leq 1$ is fine. Long story short, this result is the last component to prove some sample complexity lower bound, and argue that a sample size $n\gtrsim 1/(d\gamma^4)$ is necessary. So if $n$ is larger, the job is done already.
@ClementC. So what is the actual game? Just to show that $\psi(\gamma,d,n)$ is small if $n\gamma^4 d$ is small or it is subtler than that?
It's not subtler than that, that's indeed the goal :) @fedja
Partial progress: here is a proof for the analogous "Gaussian case" (i.e., $Z\sim N(\mathbf{0}_d, \mathbf{I}_d)$) and $n=2$. I was hoping to get some inspiration for it to handle the "Rademacher" $Z$ case I am interested in, though it doesn't seem to be helpful after all...
The goal is to bound an analogous quantity as (1), which for $n=2$ is
$$
\psi(\gamma,d,2) = \frac{e^{-\gamma^2 d}}{(1-2\gamma^2)^{d/2}}-2 + e^{\gamma^2d}\mathbb{E}_{XY}\left[ {\frac{\mathbb{E}_{Z}[ (1+\gamma\langle X,{Z}\rangle)(1+\gamma\langle Y,{Z}\rangle) ]^2}{ e^{\frac{\gamma^2}{2}\lVert X+Y \rVert_2^2} } }\right]
$$
(the denominator changes because it comes from $\mathbb{E}_Z[e^{\gamma \langle\sum_{j=1}^n X^{(j)},Z\rangle}]$, and that expression changes between the Gaussian and Rademacher cases. Similarly for why the additive "-1" is now a less nice expression.)
We can expand and compute the numerator (before squaring) as
\begin{align*}
\mathbb{E}_{Z}\left[(1+\gamma\langle X, Z\rangle)(1+\gamma\langle Y, Z\rangle) \right]
= \mathbb{E}_{Z}\left[1+\gamma\langle{X+Y,Z}\rangle + \gamma^2\langle X, Z\rangle\langle Y, Z\rangle \right]
= 1+ \gamma^2\langle X, Y\rangle
\end{align*}
since $Z$ is Gaussian. From there, observing that $V := \frac{\lVert X+Y\rVert_2^2}{2}$ is a $\chi^2(d)$ r.v., we get
\begin{align*}
e^{-\gamma^2d }&(1+\psi(d ,\gamma,2))\\
&= \mathbb{E}_{XY}[{ e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} (1+ \gamma^2\langle X, Y\rangle)^2 }]\\
&= \mathbb{E}_{V}[{ e^{-\gamma^2 V} }] + 2\gamma^2 \mathbb{E}_{XY}[{\langle X, Y\rangle e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }] + \gamma^4\mathbb{E}_{XY}[{\langle X, Y\rangle^2e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }] \\
&\leq \mathbb{E}_{V}[{ e^{-\gamma^2 V} }] + 2\gamma^2 \mathbb{E}_{XY}[{\langle X, Y\rangle e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }] + \gamma^4\mathbb{E}_{XY}[{\langle X, Y\rangle^2}] \\
&= \frac{1}{(1+2\gamma^2)^{d /2}} + \gamma^2 \mathbb{E}_{XY}[{(\lVert X+Y\rVert_2^2 - \lVert X\rVert_2^2 - \lVert Y\rVert_2^2) e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }] + d \gamma^4\,.
\end{align*}
The second term can be computed explicitly. Indeed, we have
\begin{align*}
\mathbb{E}_{XY}[{\lVert X+Y\rVert_2^2e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }]
&= 2 \mathbb{E}_{V}[{V e^{-\gamma^2V} }] = \frac{2d }{(1+2\gamma^2)^{1+d /2}} \\
\mathbb{E}_{XY}[{\lVert X\rVert_2^2e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }]
&= \mathbb{E}_{X}[{\lVert X\rVert_2^2 \mathbb{E}_{Y}[{ e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }] }]
= \mathbb{E}_{X}[{\lVert X\rVert_2^2 \prod_{i=1}^d \mathbb{E}_{Y}{ e^{-\frac{\gamma^2}{2}(X_i+Y_i)^2} } }]\\
&= {(1+\gamma^2)^{-d/2}}\mathbb{E}_{X}[{\lVert X\rVert_2^2e^{-\frac{\gamma^2}{2(1+\gamma^2)}\lVert X\rVert_2^2} }]
= \frac{d (1+\gamma^2)}{(1+2\gamma^2)^{1+d /2}}
\end{align*}
from which the second term equals
\begin{align*}
\mathbb{E}_{XY}[{(\lVert X+Y\rVert_2^2 - \lVert X\rVert_2^2 - \lVert Y\rVert_2^2) e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }]
&= - \frac{2d \gamma^2}{(1+2\gamma^2)^{1+d /2}}\,.
\end{align*}
Overall, we get
\begin{align*}
e^{-\gamma^2d }\left(2-\frac{e^{-\gamma^2 d}}{(1-2\gamma^2)^{d/2}}+\psi(d ,\gamma,2)\right)
&\leq \frac{1}{(1+2\gamma^2)^{d /2}}\left(1- \frac{2d \gamma^4}{1+2\gamma^2}\right) + d \gamma^4
\leq \frac{1}{(1+2\gamma^2)^{d /2}} + d \gamma^4\,,
\end{align*}
from which
\begin{align*}
\psi(d ,\gamma,2)
= O( d \gamma^4 )
\end{align*}
as long as $\gamma^2d = O(1)$.
|
2025-03-21T14:48:29.926215
| 2020-02-23T03:12:02 |
353355
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Bas Spitters",
"Ingo Blechschmidt",
"Martin Brandenburg",
"Tim Campion",
"Valery Isaev",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/353355"
}
|
Stack Exchange
|
Tight apartness relations in toposes
A tight apartness relation on a set is a binary relation $\#$ such that the following conditions hold:
$x = y$ if and only if $\neg (x \# y)$.
If $x \# y$, then $y \# x$.
If $x \# z$, then either $x \# y$ or $y \# z$ for every $y$.
I want to understand this notion better. Classically, it is completely trivial. Thus, it makes sense to look at it in various toposes. I tried to use the Kripke-Joyal semantics to get the external interpretation of an arbitrary object of a topos with a tight apartness relation, but it seems that it does not give anything particularly interesting in general. Thus, I've got the following question:
Question: What are examples of objects in toposes with a tight apartness relation which externally correspond to some interesting or useful notion?
Since constructively, a set can have more than one tight apartness relation on it, I'd like to see examples of an object with two different tight apartness relations which both have interesting interpretations.
Edit: There are several "generic" examples of objects with tight apartness relations, i.e., objects that can be defined in every topos (e.g., Dedekind reals). I'm particularly interested in "non-generic" examples, i.e., objects that can be constructed only in a specific topos.
I'm a little out of my depth here, but I think the first bullet point tells you that in a topos, any object admitting an apartness relation must have decidable equality. I think the first two toposes I tend to think of besides $Set$ are $sSet$ and sheaves on a topological space $X$. In $sSet$, I think maybe only discrete objects have decidable equality? And in $Sh(X)$, maybe it's just coproducts of representables on clopen sets?
The relation with the Hausdorff topology may be helpful.
@TimCampion It does not imply decidability of equality, it only implies that it is $\neg \neg$-separated. There are plenty of examples of objects with an apartness relation and without decidable equality. E.g., functions $\mathbb{N} \to \mathbb{N}$ and Dedekind reals.
@ValeryIsaev: Can you clarify? The condition "$\forall x,y : M. \neg\neg(x = y) \Rightarrow x = y$", interpreted in the internal language, is exactly the condition for the object $M$ to be separated with respect to the $\neg\neg$-topology.
@ValeryIsaev Surely you're right. Just to clarify (since it's the "tight"-ness part that has me confused more so than the "apartness" part), you're saying that in a topos, $\mathbb N^{\mathbb N}$ and $\mathbb R$ have tight apartness relations, right? BTW is there a name for objects $X$ with $\neg\neg$-dense diagonal? I didn't realize there was a condition like this on the equality relation intermediate between decidability and non-decidability...
@IngoBlechschmidt Yes, you're right. I just was confused. I deleted my comment.
@TimCampion Yes, they both carry a tight apartness relation. Also, my second comment was completely wrong. First, the condition "$\neg \neg x = y \implies x = y$" is equivalent to the diagonal being $\neg \neg$-closed and not $\neg \neg$-dense. Second, as Ingo pointed out, it is actually equivalent to the object being $\neg \neg$-separated.
@TimCampion I'm asking about tight apartness relations because every apartness relation on $X$ determine a tight one on a quotient of $X$. For example, the relation "$(x - y)$ is a unit" on a local ring is an apartness relation. The corresponding ideal is (the unique) maximal ideal consisting of non-units, the quotient is a Heyting field, and the tight apartness relation on this field is the same as before. Thus, I believe that, to understand apartness relations, it is enough to understand tight ones.
I'm not precisely sure what you're looking for. Here is an example for the external interpretation of an apartness relation:
Recall that the object of Dedekind reals $\mathbb{R}$ in a sheaf topos $\mathrm{Sh}(X)$ is the sheaf $\mathcal{C}$ of continuous (Dedekind-)real-valued functions on $X$.
The apartness relation on $\mathbb{R}$, defined by $x \mathrel{\#} y \Longleftrightarrow x - y \text{ is invertible}$, is then the following subsheaf $\mathcal{E}$ of $\mathcal{C} \times \mathcal{C}$:
$$ \mathcal{E}(U) = \{ (f,g) \,|\, \text{for all $x \in U$: $f(x) - g(x) \in \mathbb{R}$ is invertible} \} $$
I'd like to see more examples like this one, but I think even more interesting are non-generic examples. That is, objects with tight apartness relations that can be constructed only in some particular topos (here, $\mathbb{R}$ can be constructed in any topos).
@ValeryIsaev I think it would be a good idea to include this requirement into your question.
|
2025-03-21T14:48:29.926651
| 2020-02-23T04:17:32 |
353360
|
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"Alexandre Eremenko",
"M. Rahmat",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353360"
}
|
Stack Exchange
|
Extension of subharmonic function: can someone explain the details?
In this paper we have the following situation on page 60. $E$ is a compact subset of $\mathbb{R}^\tau\cup\{\infty\}$ (one point compactification) for $\tau\geq2$, $M_0$ is a point in the boundary of $E$ and $u$ is subharmonic on an open neighborhood $V$ of $M_0$. M. Brelot says that we can extend $u$ to a function that is subharmonic on a neighborhood of$E$, that he steel calls $u$, by "modifying properly $u$ outside $V$". Can someone explains the detail of this extension?
By the way a function $v$ on an open set $W$ of $\mathbb{R}^\tau\cup\{\infty\}$ is called subharmonic if: either $W$ does not contain $\infty$ and in this case $v$ is subharmonic on $W$ in the usual sense, or $W$ contains $\infty$ and then 1) $v$ is upper semicontinuous at $\infty$ and 2) $v(\infty)\leq$ the mean integral value of $u$ over any ball $B$ that is relatively compact in $W$.
Recall that the extension of a subharmonic function to $\mathbb{R}^\tau$ is classic. If $u$ is subharmonic on a neighborhood of $\overline{V}$ (closure), where $V$ is a bounded open set in $\mathbb{R}^\tau$ such that $\mathbb{R}^\tau\setminus\overline{V}$ is connected, then there exists a subharmonic function $\bar{u}$ on $\mathbb{R}^\tau$ that coincides with $u$ on $\overline{V}$ (see Armitage and Gardiner, "classical potential theory", pg192).
But Brelot extends $u$ to an open set of $\mathbb{R}^\tau\cup\{\infty\}$ containg $E$. How??
If $E$ does not contain $\infty$, you know how to extend to $R^n$. If $E$ contains $\infty$, suppose that $x_0\not\in E$, and perform the Kelvin transform which sends $x_0$ to $\infty$. If $E=R^n\cup{\infty}$ then there is no boundary point and no argument is needed.
Consider the case of $E$ containing $\infty$. Suppose the ball $B(x_0,r)\subset \complement E\cap V$ ($\complement E$=complement of $E$) and $u^$ is the Kelvin transfrom of $u$ relative to the sphere $\partial B(x_0,r)$. If I am following correctely, $u^$ is subharmonic on a neighborhood of $E$, but the restriction of $u^*$ to $V$ does not coincide with $u$. I need the restriction of the extension to coincide with $u$ on $V$ or $\overline{V}\cap E$ (assume $u$ is subharmonic on a neighborhood of $\overline{V}$) .
If I understand correctly (unfortunately I do not speak French), Brelot refers here to one of his previous papers: Sur le rôle du point à l'infini dans la théorie des fonctions harmoniques, Ann. Éc. Norm. sup., t. 61, 1944. I did not follow this reference, but I guess the argument is not much different from the classical one.
Yes, he refers to part 20, "potential representation", of the paper you cited. But I only found there a generalization of the classic result that a subharmonic function can be locally represented as the sum of a potential and a harmonic function. I didn't get what does it have to do with the extension of subharmonic functions...
|
2025-03-21T14:48:29.926871
| 2020-02-23T06:11:59 |
353362
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Tim Campion",
"https://mathoverflow.net/users/2362"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/353362"
}
|
Stack Exchange
|
Which lattices are quotients of finite powerset lattices?
Let $S$ be a finite set, and let $2^S$ be its powerset, regarded as a lattice. Let $L$ be a quotient (in the category of lattices and maps which preserve $\top,\bot,\wedge,\vee$) of $S$. What can we say about $L$?
In fact, what I'd really like to know is: which finite semilattices are retracts (via $\bot,\vee$-preserving maps) of finite powerset lattices. I think the two questions are equivalent via a short-but-not-immediate argument.
Note that if $P$ is an arbitrary finite poset, then the lattice $2^P$ of poset maps $P \to 2$ is an example of such a lattice.
The class of finite powerset lattices is closed under quotients, up to isomorphism. That is, the quotients are exactly the lattice reducts of finite Boolean algebras.
In particular, any quotient $L$ of $2^S$ has to be a bounded distributive lattice, as the class of distributive lattices is a variety. Moreover, if $x\in L$ is an image of a set $A\subseteq S$, then the image of $S\smallsetminus A$ is an element $y$ such that $x\lor y=\top$ and $x\land y=\bot$. Thus, $L$ is complemented, i.e., it is a Boolean algebra, and as such it is isomorphic to $2^{S'}$ for some set $S'$.
Tim-
Emil answered the question you asked, but since you wrote ``what I'd really like to know is: which finite semilattices are retracts (via $\bot$, $\vee$-preserving maps) of finite powerset lattices'' let me add to his answer.
The $2$-element semilattice is injective in the class of semilattices.
The class of injectives is closed under products and retracts.
Up to isomorphism, the powers of the $2$-element semilattice are the power-set semilattices. Hence retracts of power-set semilattices must be injective.
Conversely, since the $2$-element semilattice is the only subdirectly irreducible semilattice, every semilattice is embeddable in some power $2^S$. And since an injective is a retract of any extension, it follows that every injective arises as a retract of some $2^S$,
Thus, the retracts of the power-set semilattices, $2^S$, are exactly the injective semilattices.
Theorem 2.8 of
The Category of Semilattices
ALFRED HORN and NAOKI KIMURA
Algebra universalis 1 (1971), 26-38.
proves that a (meet-)semilattice is injective iff it is complete and satisfies an infinite distributive law, namely that the meet distributes over infinite joins.
It follows that a finite semilattice is a retract of a power-set semilattice iff it is the semilattice reduct of a finite distributive lattice. (This shows that the examples in your last struckout paragraph exhaust all examples.)
Thanks! I had just convinced myself of this using a different argument, but I'm very glad to know that there is an injectivity characterization like this, which makes it apparent.
|
2025-03-21T14:48:29.927058
| 2020-02-23T07:39:34 |
353363
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/353363"
}
|
Stack Exchange
|
A closed subspace generated by open $F_{\sigma}$ sets of $K$ in $C(K)^{**}$
Let $K$ be a compact Hausdorff space. For a bounded Borel measurable function $f$ on $K$, we define $\phi_{f}\in C(K)^{**}$ by $\phi_{f}(\mu)=\int_{K}fd\mu$ for all $\mu\in C(K)^{*}$. It is easy to see that $\|\phi_{f}\|=\|f\|_{\infty}$. Thus the space of all bounded Borel measurable functions on $K$ can be considered as a closed subspace of $C(K)^{**}$. We let $Z$ be the norm closed subspace of $C(K)^{**}$ generated by $\{\chi_{U}:U$ open $F_{\sigma}$ sets in $K\}$. Given a sequence $(U_{n})_{n}$ of disjoint open $F_{\sigma}$ sets in $K$ and a bounded scalar sequence $(a_{n})_{n}$, we define $g(t)=a_{n}$ for $t\in U_{n}$ and $g(t)=0$ for $t\notin \cup_{n}U_{n}.$ Then $g$ is a bounded Borel measurable function. It seems clear that $g$ belongs to $Z$. But I am not sure.
Question. Does $g$ belong to $Z$?
Let $\epsilon > 0$. Since $a_n$ is a bounded sequence, by compactness we can find a finite set of scalars $b_1, \dots, b_m$ such that for every $a_n$ there exists a $b_{k_n}$ with $|a_n - b_{k_n}| \le \epsilon$. Now consider the function $h = \sum_{n=1}^\infty b_{k_n} \chi_{U_n}$, that is, $h(t) = b_{k_n}$ for $t \in U_n$. Clearly $\|\phi_g - \phi_h\| = \|g-h\|_{\infty} \le \epsilon$. Now set $V_k = \bigcup_{k_n = k} U_n$. Then $V_k$ is an open $F_\sigma$ since it is a countable union of open $F_\sigma$s, and we can write $h = \sum_{k=1}^m b_k \chi_{V_k}$. Thus $h \in Z$, and since $\epsilon$ was arbitrary and $Z$ is closed, we have $g \in Z$ also.
|
2025-03-21T14:48:29.927181
| 2020-02-23T08:49:11 |
353366
|
{
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"Kevin Casto",
"Lira",
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"YCor",
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"user44191"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/353366"
}
|
Stack Exchange
|
Homogeneous polynomial map inducing diffeomorphism of $\mathbb{R}^3\setminus\{0\}$
Assume that $f(x) = (p_1(x),p_2(x),p_3(x))$ is a homogeneous polynomial inducing a diffeomorphic mapping of $\mathbb{R}^3\setminus\{0\}$ onto itself. Homegeneous means $f(tx)=t^n f(x)$, for $t>0$ and some $n\in \mathbb{N}$. Why $n$ should be 1?
$f(x)=(x_1^3,x_2^3,x_3^3)$ is counterexample.
@YCor. This is not a diffeomorphism of $\mathbb{R}^3\setminus{0}$ onto itself. Its Jacobian is $3^3x_1^2 x_2^2 x_3^2=0$ not only on $(0,0,0)$.
Thanks, you're right... sorry! I'm leaving it for other users...
$f(x)= |x|^2 x$ is a counterexample?
That's not a polynomial.
@ZachTeitler I think it is - $|x|^2 = x_1^2 + x_2^2 + x_3^2$ is a polynomial.
Oh right! Thanks.
@KevinCasto: Isn't the Jacobian $3(x^2+y^2+z^2)^3$?
@MateuszKwaśnicki oops you're right!
Note: the inverse diffeomorphism is given by $x\mapsto |x|^{-2/3}x$.
|
2025-03-21T14:48:29.927432
| 2020-02-23T11:09:26 |
353374
|
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"Bill Johnson",
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"https://mathoverflow.net/users/2554",
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"url": "https://mathoverflow.net/questions/353374"
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|
Stack Exchange
|
Banach lattices $X$ for which $L_p(\mu)\subset X$ or $X\subset L_q(\mu)$
It is well known (see vol. II of Lindenstrauss and Tzafriri's book) that an order continuous Banach lattice $X$ with a weak unit admits a representation as a (in general not closed) ideal of $L_1(\mu)$ for some probability measure $\mu$ so that $L_\infty(\mu)\subset X\subset L_1(\mu)$ with continuous embeddings.
I am interested in conditions on $X$ implying that $L_p(\mu)\subset X$ for some $p>1$, or $X\subset L_q(\mu)$ for some $q<\infty$.
The discussion in LT2 on Boyd indices is not sufficient for you? Note the remark after the proof of Proposition 2.b.3 that to prove the proposition you only need the estimates on the fundamental function $|1_{(0,t)}|_X$. I think having the needed estimates on the fundamental function is almost equivalent to the containments you want.
Proposition 2.b.3 and the subsequent remarks are useful for me.
|
2025-03-21T14:48:29.927520
| 2020-02-23T11:45:50 |
353375
|
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"Henri Johnston",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353375"
}
|
Stack Exchange
|
$C_4\times C_2 : C_2$: what does this mean?
I am reading this paper where the object $C_4\times C_2 : C_2$ is used as a group structure. I know that $C_n$ is a cyclic group but don't know what kind of operation between groups is identified by the symbol "$:$". Does anyone know about that? Thanks to everyone for the help.
This paper : https://www.worldscientific.com/doi/abs/10.1142/S0219498816501759
As noted in the answer below, it's semidirect product, but does not specify which one. This list of groups of order 16 might be helpful: https://people.maths.bris.ac.uk/~matyd/GroupNames/index.html#order16 The "aliases" feature is particularly useful.
@Henri Johnston Thanks for the link.
The colon means "semidirect product", but it does not specify which semidirect product. This notation is a concise shorthand that gives important structural information without necessarily uniquely specifying the group. You can read more about similar notation conventions in the introduction to the ATLAS of finite groups.
Thanks for your answer.
|
2025-03-21T14:48:29.927629
| 2020-02-23T12:28:52 |
353377
|
{
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"authors": [
"Charlie Frohman",
"Danny Ruberman",
"Moishe Kohan",
"https://mathoverflow.net/users/3460",
"https://mathoverflow.net/users/39654",
"https://mathoverflow.net/users/4304"
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|
Stack Exchange
|
Does every $SL_2\mathbb{C}$ representation of a closed oriented surface extend over a compact oriented three-manifold?
Let $F$ be a compact oriented surface and $\rho:\pi_1(F)\rightarrow SL_2\mathbb{C}$ be a representation. Does there exist a compact oriented three-manifold $M$ with $\partial M=F$ and a homomorphism $\tilde{\rho}:\pi_1(M)\rightarrow SL_2\mathbb{C}$ so that the restriction of $\tilde{\rho}$ to $\pi_1(F)$ is equal to $\rho$?
If not is there an obstruction that allows you to identify the representations that do extend?
For instance let $BSL_2\mathbb{C}^\delta$ denote the classifying space of $SL_2\mathbb{C}$ as a discrete group. Corresponding to $\rho:\pi_1(F)\rightarrow SL_2\mathbb{C}$ is a continuous map $f:F\rightarrow BSL_2\mathbb{C}^\delta$. If $\rho$ extends over a three-manifold $M$ then the homology class represented by $f_*[F]$ is zero. Is there a computable way to detect this?
This is unlikely. The expected dimension of $X(M)$ is half of the dimension of $X(F)$, where $X$ stands for the character variety. Since there are only countably many compact 3-manifolds $M$, "most" points of $X(F)$ probably not come from restrictions.
Hi Charlie--It seems to me that you are asking if there are non-trivial classes in $H_2(BG^\delta)$, or perhaps how to detect such classes. You might like to have a look at Milnor's paper, On the homology of Lie groups made discrete. He cites an argument in a paper of Alperin-Dennis (due to Mather) for the case of G=SL(2,R) that might be relevant.
Here is an argument that "most" points in the $SL(2, {\mathbb C})$-character variety $X(F)$ of the surface $F$ do not correspond to representations extendible to 3-manifold groups (as in the question).
Let $M$ be a compact oriented 3-manifold with $\partial M=F$. We then have the "restriction morphism" of $SL(2, {\mathbb C})$-character varieties
$$
r: X(M)\to X(F).
$$
The image of $r(X(M))$ is "formally Lagrangian" (more precisely, Lagrangian on the scheme-theoretic level) with respect to the standard complex symplectic structure on $X(F)$, see
A. Sikora, Character varieties. Trans. Amer. Math. Soc. 364 (2012), no. 10, 5173–5208.
In particular, $\dim r(X(M))\le \frac{1}{2} \dim X(F)$. Since there are only countably many 3-manifolds $M$ as above, the union
$$
U=\bigcup_{M} r(X(M)) \subset X(F)
$$
has empty interior (in the Euclidean topology). Thus, "most" points in $X(F)$ do not belong to $U$. I do not know how to detect non-membership in this union algorithmically. Since you are working over the complex numbers, you have to specify what computability even means. For instance, you can restrict to $\overline{{\mathbb Q}}$-points of the character variety (i.e. equivalence classes of representations to $SL(2, \overline{{\mathbb Q}})$); then at least one can use the classical notion of computability and your question is well-defined in this setting. (There is a silly algorithm which terminates for points in $U(\overline{{\mathbb Q}})$.). I do not even know if the membership problem in $U(\overline{{\mathbb Q}})$ is decidable.
PS. This entire discussion feels related to the proof of Theorem 1.3 in
N. Dunfield, W. Thurston, Finite covers of random 3-manifolds. Invent. Math. 166 (2006), no. 3, 457–521.
Edit. Here is one way to find explicit examples of non-extendible representations $\rho$, i.e. such that $[\rho]$ does not belong to $U$ (motivated by Ian Agol's answer in the case of genus $1$). I will use the fact that the variety $X(F)$ is ${\mathbb Q}$-rational, see for instance Theorem 2 in
A. Rapinchuk, V. Benyash-Krivetz, V. Chernousov, Representation varieties of the fundamental groups of compact orientable surfaces. Israel J. Math. 93 (1996), 29–71.
In other words, there exists a birational isomorphism defined over ${\mathbb Q}$, $f: X(F)\to {\mathbb C}^{6g-6}$.
Hence, instead of $X(F)$ we can essentially work in ${\mathbb C}^{6g-6}$ (with its standard rational structure).
Now, take a point $p=(z_1,...,z_{6g-6})$ in ${\mathbb C}^{6g-6}$ whose coordinates generate a field of transcendence degree $>3g-3$. (Such $p$ necessarily belongs to the image of $f$ and $f^{-1}(p)$ is a singleton.) One can find such tuples $p$, for instance, using the Lindemann–Weierstrass theorem. Then $[\rho]=f^{-1}(p)$ does not lie in $U$.
Sorry, I am pretty unsophisticated about ``computablity''. Where can I learn about the algorithm for coefficients in $\overline{Q}$?
@CharlieFrohman: The algorithm, as I said, is silly and totally non-practical: Enumerate all triangulated compact 3-manifolds with boundary $F$. For each of these manifolds, we have a polynomial morphism $r$ as above. For a morphism of affine varieties $r: Y\to X$, there is an algorithm for computation of defining equations and inequalities of $r(Y)$. (I will find a reference if you like.) Then check these on the given point in $X(F)$. If $[\rho]\in r(X(M))$ for some $M$, then this algorithm will terminate and find the manifold $M$.
In addition to Moishe Kohan's geometric argument, there's also a
bordism-theoretic proof.
$\newcommand{\BDel}{B\mathrm{SL}_2(\mathbb C)^\delta}$
Let $\Omega_*^{\mathrm{SO}}(-)$ denote oriented bordism as a generalized homology theory. Your question is
equivalent to asking whether $\Omega_2^{\mathrm{SO}}(\BDel) = 0$.
We can compute this with the Atiyah-Hirzebruch spectral sequence, which has signature
$$
E^2_{p,q} = H_p(\BDel, \Omega_q^{\mathrm{SO}}(\mathrm{pt})) \Longrightarrow \Omega^{\mathrm{SO}}_{p+q}(\BDel).
$$
$\Omega_q^{\mathrm{SO}}(\mathrm{pt}) = 0$ for $q = 1,2,3$, so in the range $p+q < 4$, this spectral sequence
collapses, implying $\Omega_2^{\mathrm{SO}}(\BDel) \cong H_2(\BDel; \mathbb Z)$. Now, as suggested by Danny
Ruberman's
comment,
Milnor's “On the homology of Lie groups made discrete”
points out that $H_2(\BDel; \mathbb Z)$ surjects onto an uncountable $\mathbb Q$-vector space, hence is in particular
nontrivial, and therefore not every $\mathrm{SL}_2(\mathbb C)$-representation of a closed, oriented surface extends
to a compact $3$-manifold.
Unfortunately, this approach is difficult to make explicit for a given representation of a surface group without a
better understanding of the homology of $\mathrm{SL}_2(\mathbb C)$ as a discrete group.
Thanks! I need to read this Milnor paper.
This is an extended comment on the algorithmic question.
As Moshe points out, the lack of realization as a boundary follows from the Baire category theorem. On the other hand, how does one recognize when an element is not in this infinite countable union of subspaces?
Let's consider the genus 1 case $F=T^2$. Then a rep $\rho:\pi_1(F)\to SL_2(\mathbb{C})$ is determined by two eigenvalues of generators $(\mu, \lambda)$ (assuming the representation is not unipotent). In turn if $F=\partial M$, the boundary of a 3-manifold, then there is an associated $A$-polynomial $A(x,y) \in \mathbb{Z}[x,y,x^{-1},y^{-1}]$ such that $A(\mu,\lambda)=0$. I suspect that $[\rho]=0\in H_2(SL_2(\mathbb{C})^\delta)$ iff $[\rho]$ extends to a representation of a 3-manifold, but I haven't checked this. In any case, one sees that $\mu, \lambda$ are algebraically related in the bounding case. However conversely, if $\mu,\lambda$ satisfy an algebraic relation, it's not clear to me that this implies that $[\rho]=0$, since in general A-polynomials satisfy some non-trivial conditions.
I suspect there could be a formulation in terms of algebraic K-theory, but I don't know enough about this.
One may also ask if $H_2(SL_2(\mathbb{C})^\delta)$ is generated by representations of $T^2$? I suspect this might be true. It's not hard to see that a representation of a closed surface of genus $>2$ is cobordant to a sum of representations of genus 2 surfaces (since the commutator map in $SL_2(\mathbb{C})$ is onto). Then I think that genus 2 reps. may be cobordant to a pair of genus 1 reps. (at least the numerology works out, but I haven't checked it). Then one could ask for when a sum of genus 1 reps. is homologically trivial? In turn, this should be realized by a zero of an A-variety. But I'm not sure how one recognizes such points.
|
2025-03-21T14:48:29.928134
| 2020-02-23T15:36:19 |
353386
|
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|
Stack Exchange
|
Geometry of Level sets of elliptic polynomials in two real variables
Updated:
A polynomial $P(x,y)\in \mathbb{R}[x,y]$ is called an elliptic polynomial if its last homogeneous part does not vanish on $\mathbb{R}^2\setminus\{0\}$.The two answers to this post provide a proof for the following theorem:
Theorem: If $p$ is an elliptic polynomial whose last homogeneous part is positive definitive, then for $c$ sufficiently large
, $p^{-1}(c)$ is a simple closed curve. Moreover if the centroid of interior of $p^{-1}(c)$ is denoted by $e_c$ then $e_c$ is convergent as $c$ goes to $+\infty$. The limit $\lim_{c\to \infty} e_c$ can be written in terms of coefficients of $p$. If we drop the ellipticity condition then this convergence result is not necessarily true.
The previous version of the post:
Is there a polynomial function $P:\mathbb{R}^2 \to \mathbb{R}$ with the following property?
For sufficiently large $c>0$, $P^{-1}(c)$ is a simple closed curve $\gamma_c$, homeomorphic to $S^1$, but as $c$ goes to $+\infty$. the centroid $e_c$ of the interior of $\gamma_c$
does not converge to any point of $\mathbb{R}^2$.
Motivation: The answer is negative if we consider this question for polynomials $p:\mathbb{R} \to \mathbb{R}$ whose eventual level sets are $2$-pointed set, i.e. $S^0$.(Namely a polynomial of even degree). The motivation comes from line -3, item III, page 4 of Taghavi - On periodic solutions of Liénard equations, which can be generalized to every even degree polynomial with one variable.
Concerning homogeneous polynomials: Let $P(x,y)=\sum_{j=0}^n a_j x^j y^{n-j}$ be such a polynomial, of degree $n$ such that $C:=P^{-1}(\{c\})$ is a simple closed curve for all large enough $c>0$.
If $n$ is odd, then every line through the origin will have at most one point of intersection with $C$. So, then $C$ cannot be a simple closed curve for any real $c$ -- because every line through any point interior to a simple closed curve must intersect the curve in at least two points.
It remains to consider the case when $n$ is even. Then $C$ is symmetric about the origin, and hence so is the interior of $C$. Then the centroid of the interior is the origin, and it does not depend on the level $c$.
Consider now the case of an elliptic polynomial
\begin{equation*}
P(z)=P(x,y)=\sum_{j=0}^n a_j x^j y^{n-j}+\sum_{j=0}^{n-1}b_j x^j y^{n-1-j}+K|z|^{n-2}
\end{equation*}
of (necessarily even) degree $n$,
where $z:=(x,y)$ and $K=O(1)$ (as $|z|\to\infty$). The ellipticity here is understood as the following condition:
\begin{equation*}
\min_{|z|=1}\sum_{j=0}^n a_j x^j y^{n-j}>0.
\end{equation*}
For any $d_*\in(0,1)$ and any real $D>0$, let $\mathcal P_{n,d_*,D}$ denote the set of all polynomials $p(x)=\sum_{j=0}^n d_j x^j$ such that $d_n\ge d_*$ and $\sum_{j=0}^n|d_j|\le D$.
Then it is not hard to see that there is a real $c_*(n,d_*,D)>0$, depending only on $n,d_*,D$, such that
for any polynomial $p(x)=\sum_{j=0}^n d_j x^j$ in $\mathcal P_{n,d_*,D}$ and for all real $c\ge c_*(n,d_*,D)$ the equation $p(x)=c$ has exactly two roots $x_\pm:=x_\pm(c)$ such that $x_-<0<x_+$ and, moreover,
\begin{equation*}
x_\pm=\pm\Big(\frac c{d_n}\Big)^{1/n}-(1+o(1))\frac{d_{n-1}}{nd_n} \tag{1}
\end{equation*}
uniformly over all polynomials $p(x)=\sum_{j=0}^n d_j x^j$ in $\mathcal P_{n,d_*,D}$; here and in the sequel the asymptotic relations are for
$$c\to\infty,$$
unless otherwise specified.
This uniformity can be obtained by refining this reasoning.
Moreover, without loss of generality (wlog),
\begin{equation*}
\text{for all $p\in\mathcal P_{n,d_*,D}$ and all real $c\ge c_*(n,d_*,D)$ we have $p'(x_\pm)\ne0$.} \tag{1.5}
\end{equation*}
Indeed, because (1) holds uniformly over all $p\in\mathcal P_{n,d_*,D}$, wlog
\begin{equation*}
|x_\pm|\ge\Big(\frac cD\Big)^{1/n}-2\frac D{nd_*}\to\infty, \tag{1.6}
\end{equation*}
so that $|x_\pm|\to\infty$ uniformly over all $p\in\mathcal P_{n,d_*,D}$.
Also, taking any polynomial $p(x)=\sum_{j=0}^n d_j x^j$ in $\mathcal P_{n,d_*,D}$ and writing $p'(x)=\sum_{j=1}^n d_j jx^{j-1}$, we see that for $|x|\ge1$
\begin{equation*}
\frac{|p'(x)|}{|x|^{n-1}}\ge nd_n-\sum_{j=1}^{n-1} |d_j| j|x|^{j-n}
\ge nd_*-n D |x|^{-1}\underset{x\to\infty}\longrightarrow nd_*>0.
\end{equation*}
So, by (1.6), wlog (1.5) holds indeed.
Let us now turn back to the elliptic polynomial $P(x,y)$. For each real $t$ consider the polynomial
\begin{equation*}
p_t(r):=P(r\cos t,r\sin t).
\end{equation*}
By the ellipticity of the polynomial $P(x,y)$, there exist $d_*\in(0,1)$ and a real $D>0$ such that $p_t\in\mathcal P_{n,d_*,D}$ for all real $t$. Take now any real $c\ge c_*(n,d_*,D)$. Then, by the paragraph right above, for each real $t$ the equation $p_t(r)=c$ has exactly two roots $r_\pm(t):=r_\pm(c;t)$ such that
$r_-(t)<0<r_+(t)$ and, moreover,
\begin{equation*}
r_\pm(t)=\pm\Big(\frac c{a(t)}\Big)^{1/n}-(1+o(1))\frac{b(t)}{na(t)}
\end{equation*}
uniformly in real $t$,
where
\begin{equation*}
a(t):=\sum_{j=0}^n a_j \cos^jt\, \sin^{n-j}t,\quad
b(t):=\sum_{j=0}^{n-1} b_j \cos^jt\, \sin^{n-1-j}t.
\end{equation*}
Moreover, $\frac{dp_t(r)}{dr}|_{r=r_\pm(t)}\ne0$. So, by the implicit function theorem, the functions $r_\pm$ are continuous (in fact, infinitely smooth).
Also, the functions $r_\pm$ are periodic with period $2\pi$, since for each real $t$ we have $p_{t+2\pi}=p_t$ and the values $r_\pm(t)$ of the the functions $r_\pm$ at $t$ are uniquely determined by the polynomial $p_t$. Furthermore, for all real $r$ and $t$ we have $p_{t+\pi}(r)=p_t(-r)$, which implies $r_+(t+\pi)=-r_-(t)$. So, letting
$$z_\pm(t):=r_\pm(t)(\cos t,\sin t),$$
we see that $z_\pm(t+2\pi)=z_\pm(t)$ and
$z_+(t)=z_-(t-\pi)$ for all real $t$. So,
the $c$-level curve of $P(x,y)$ is
\begin{align*}
C=P^{-1}(\{c\})&=\{z_+(t)\colon t\in\mathbb R\}\cup\{z_-(t)\colon t\in\mathbb R\} \\
&=\{z_+(t)\colon t\in[0,2\pi)\}\cup\{z_-(t)\colon t\in[0,2\pi)\} \\
&=\{z_+(t)\colon t\in[0,2\pi)\} \\
&=\{z_+(t)\colon t\in[0,\pi)\}\cup\{z_-(t-\pi)\colon t\in[\pi,2\pi)\} \\
&=\{z(t)\colon t\in[0,2\pi)\},
\end{align*}
where
\begin{equation*}
z(t):=R(t)(\cos t,\sin t), \quad
R(t):=
\begin{cases}
r_+(t)>0&\text{ for }t\in[0,\pi],\\
|r_-(t-\pi)|>0&\text{ for }t\in[\pi,2\pi].
\end{cases}
\end{equation*}
So, the level curve $C$ is closed and simple, and its interior is
\begin{equation*}
I(c):=\{r\,(\cos t,\sin t)\colon0\le r<R(t)\}.
\end{equation*}
The main idea for the elliptic polynomial case is to consider, for all real $c\ge c_*(n,d_*,D)$, the two opposite infinitesimal sectors of the interior $I(c)$ of the simple closed curve $C=P^{-1}(\{c\})$ between the rays $t$ and $t+dt$ and between the rays $t+\pi$ and $t+\pi+dt$, where $t$ is the polar angle in the interval $[0,\pi)$. The centroid of the union of these two sectors of $I(c)$ is at (signed) distance
\begin{equation*}
d(t)\sim \frac23\,\Big(r_+(t)\frac{|r_+(t)|^2}{|r_+(t)|^2+|r_-(t)|^2}
+r_-(t)\frac{|r_-(t)|^2}{|r_+(t)|^2+|r_-(t)|^2}\Big) \tag{2}
\end{equation*}
from the origin.
Formula (2) follows because (i) the centroid of an infinitesimal sector of radius $r>0$ between the rays $t$ and $t+dt$ is at distance $\frac23\,r$ from the origin, (ii) the area of this sector is $\frac12\,r^2\,dt$, and (iii) the centroid of the union of the two sectors is the weighted average of the centroids of the two sectors, with weights adding to $1$ and proportional to the areas of the sectors, and thus proportional to the squared radii of the sectors.
Simplifying (2), we get
\begin{equation*}
d(t)\sim-\frac{2b(t)}{na(t)}.
\end{equation*}
Averaging now over all the pairs of opposite infinitesimal sectors, we see that the centroid converges to
\begin{align*}
&-\int_0^\pi dt\,\frac{2b(t)}{na(t)}(\cos t,\sin t)\frac12\,\Big(\frac c{a(t)}\Big)^{2/n}
\Big/\int_0^\pi dt\,\frac12\,\Big(\frac c{a(t)}\Big)^{2/n} \\
&=-\int_0^\pi dt\,\frac{2b(t)}{na(t)}(\cos t,\sin t)\Big(\frac1{a(t)}\Big)^{2/n}
\Big/\int_0^\pi dt\,\Big(\frac1{a(t)}\Big)^{2/n}. \tag{3}
\end{align*}
I have checked this result numerically for $P(x,y)=x^4 + y^4 + 3 (x - y)^4 + y^3 + x y^2 + 10 x^2$, getting the centroid $\approx(-0.182846, -0.245149)$ for $c=10^4$ and $\approx(-0.189242,-0.25)$ for the limit (as $c\to\infty$) given by (3). From the above reasoning, one can see that the distance of the centroid from its limit is $O(1/c^{1/n})$; so, the agreement in this numerical example should be considered good, better than expected.
One may also note that in general the level sets $P^{−1}([0,c])$ will not be convex, even if $P$ is a positive elliptic homogeneous polynomial. E.g., take $P(x,y)=(x−y)^2(x+y)^2+h(x^4+y^4)$ for a small enough $h>0$. Here is the picture of this level set for $c=1$ and $h=1/10$:
Clearly, the shape of this level set does not depend on $c>0$.
This non-convexity idea can be generalized, with
$$P(x,y)=P_{k,h}(x,y)
:=\prod_{j=0}^{2k-1}\Big(x\cos\frac{\pi j}k-y\sin\frac{\pi j}k\Big)^2+h(x^{4k}+y^{4k})$$
for natural $k$ and real $h>0$. Here is the picture of the curve $P_{k,h}^{-1}(\{1\})$ for $k=5$ and $h=(3/10)^{4k}$:
Thanks for your answer. What about an elliptic polynomial?
@AliTaghavi : How do you define an elliptic polynomial?
I am not sure my terminoligy is good.(please see the next comment too. But what was in mind is the following: The last homogeneous part is positive (or negative) on punctured plane. Example $x^4+y^4+ \text{lower terms}$. In fact every polynomial whose corresponding pde is elliptic.
https://mathoverflow.net/questions/117437/the-distribution-of-roots-of-elliptic-polynomial
They are not equivalent but what I was thinking about initially is the first one.
It would be interesting if we could prove that the centroid of level set of a (somewhat) elliptic polynomial is convergent and the cordinates of limiting centroid can be expreced in terms of coefficients of polynomial. (May be in terms of coefficients of degree 2n and 2n-1 and not lower degree. This is really the case for one variable)
I have added the case of elliptic polynomials.
I have removed the "eventual" convexity condition (which was previously imposed for elliptic polynomials).
Once again I thank you very much for your answer I try to understand its details.
Please let me know where details are needed the most. Since the answer is already very long, I felt it necessary to omit some of the details.
To be honnest i have some question about your answer. But this is my first question: in this new version of your answer, how do not you need the convexity assumption? Is not possible that a ray intersect our closed curve in several points?
@AliTaghavi : The convexity is not needed because we now have the uniformity: (1) holds uniformly over all polynomials in $\mathcal P_{n,d_,D}$ if $c\ge c_(n,d_*,D)$. So, if $c>0$ is large enough, then for all $t\in[0,2\pi)$ at once the polynomials $P(r\cos t,r\sin t)-c$ in $r$ have exactly two roots, $r_\pm(t)$, satisfying condition (2a). I have now inserted the previously missing qualification "for all large enough $c>0$" into the sentence "The main idea for the elliptic polynomial case is ...".
Yes I see. Please give me a few days to underestand your interesting answer. But just a question: Do you think my assumption that for c sufficiently large the preimage of c is simple clised curve is redandant?
I really wish to understand your answer but I admit it is a little complicated. BTW what (equivalent formulation of centroid you are using? $\iint (x,y)dxdy$ divided by area?
@AliTaghavi : For all large enough $c$, the $c$-level curve of $P$ is ${R(t)(\cos t,\sin t)\colon t\in[0,2\pi)}$, where $R(t):=r_+(t)$ for $t\in[0,\pi)$ and $R(t):=r_-(t-\pi)$ for $t\in[\pi,2\pi)$. So, the level curve is closed and simple, and there is no need to additionally assume that. As for the centroid, this is the way I understand it; is there any other way?
Can you please give a reason? The cordinate (sin t, Cost) implies the convexity? Does not? Just other question why it is not possible that large level sets would be open curves? From Morse theorem and also Bezout theorem we know that for a generic polynomial p, all large level sets are homeomorphic to each other but it does not guarantee that they are closed curv, yes?
@AliTaghavi : I have added details on why the $c$-level curve for an elliptic polynomial is necessarily simple and closed for all large enough $c>0$. I have not stated or used any convexity properties.
Thank you so much!
I appreciate again your very onteresting answer. I read it several times I am finalizong my full understanding.
@AliTaghavi : I am glad you liked the answer. Please let me know if more clarifications are needed in some places.
Thank you for your comments. Is there an easy argument for "wlog we may assume "c" is a regular value that is $p'(r^+)$ and $p'(r^-)$ are non zero. You uses this fact in 2 variable setting.(Sòrry if my question is elementary).
I mean is it obvious that we may assume that c is regular value UNIFORMLY independent of choosing coefficient in $\mathcal{p} (n,d_*, D)$?
BTW do we realy need the convexity? And is it realy the case?Each line passing irigin intersect the level set at exactly 2 points but it does not necessarily imply that it is convex, right? So is the theorem which I wrote in the first paragraph completly true? Should not we remive the word convex?
@AliTaghavi : I have now added details on why without loss of generality $p'(x_\pm)\ne0$ for all $p\in\mathcal P_{n,d_,D}$ and all real $c\ge c_(n,d_*,D)$ -- this statement is now formula (1.5).
@AliTaghavi : I did use an additional convexity assumption in an earlier version of the answer, but later was able to remove it. Indeed, no such assumption is needed to obtain the limit of the centroid.
@AliTaghavi : Also, in general the level sets $P^{-1}((-\infty,c])$ will not be convex, even if $P$ is an elliptic homogeneous polynomial. E.g., take $P(x,y)=(x - y)^2 (x + y)^2 + (x^4 + y^4)/10$.
$(y-x^2)^2+x^2\phantom{aaaaaaaaaaaaaaaaaaaaa}$
Thanks for your answer. Is there a homogenous example or a polynomial whose last homogenous part os elliptic(non degenerate)?
In case of convergence of centroid to a point q, how can one write q in terms of coefficoent of the polynomial p?(as in one variable). BTW it seems that the elliptic assumption is a reasonable generalization of 1 variable case.
You actually send the centroid to infinity along the y axis. I think your example is based on choosing a volum preserving diffeomorphism as change of cordinate. But is there really a homogenuos or elliptic example?
Apart from above questions, is there an example for which the centroid is bounded but is not convergent?(very irregular behaviour)?
To be honest, before your answer I was interested in a kind of non degeneracy conditioñs on the last homogeneous part please see the comment conversations on this post
https://math.stackexchange.com/questions/3547528/every-even-degree-polynomial-is-eventually-symmetric/3551129#3551129
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2025-03-21T14:48:29.929030
| 2020-02-23T15:44:02 |
353388
|
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|
Stack Exchange
|
Solutions in primes of the equation $\,3p^2+q^2=r^2+3$
Let's consider the Diophantine equation $\,3p^2+q^2=r^2+3$.
Actually, I am interested only in the solutions represented by sets $\,(p,q,r)\,$ of prime numbers.
It's easy to prove that if $\,(p,q)\,$ is a pair of twin primes $\,(p\lt q)$, then $\,(p,q)\,$ solves the equation if $\,p\,$ is a Sophie Germain prime and $\,r\,$ represents its safe prime (that is $\,r=2p+1$).
Similarly, if $\,(p,q)\,$ is a pair of twin primes with $\,p\gt q$, then $\,(p,q)\,$ solves the equation if $\,r=2p-1$.
Therefore, here are some solutions of the given equation:
$(3,5,7)\;\;(5,7,11)\;\;(11,13,23)\;\;(29,31,59)\;\;(41,43,83)\;\;...$
$(7,5,13)\;\;(19,17,37)\;\;(31,29,61)\;\;...$
In order to find other classes of solutions, let's force $\,r=p+2q$. This constraint leads to consider only the pairs $\,(p,q)\,$ which satisfy the following condition:
$5(q^2+1)=2((p-q)^2+1)\;\;\;\;\;\;\;\;\;(*)$
A solution satisfing $\,(*)\,$ is $\,(13,5,23)$.
Other solutions are:
$(29,17,53)\;\;(41,17,73)\;\;(59,13,103)\;\;(61,17,107)\;\;(71,107,163)\;\;(79,7,137)\;\;(89,53,163)$
The previous solutions have been found observing that the equation can be written
$$3(p^2-1)=r^2-q^2$$
So, $\,72\,$ always divide $r^2-q^2$.
I ask to find other, possibly more general, classes of solutions of the given equation.
Are you sure you mean terns of primes?
Generalizing in another way from the solution $(3,5,7)$, there is also a class of solutions in which $(q,r)$ is a pair of twin primes ($q<r$). A parametric solution with $r=q+2$ is:
$$(p,q,r)=(2m+1,3m^2+3m-1,3m^2+3m+1)$$
Values of $m$ which yield solutions in primes include $1,2,9,30$ giving respectively:
$(3,5,7)\;\;(5,17,19)\;\;(19,269,271)\;\;(61,2789,2791)$
|
2025-03-21T14:48:29.929153
| 2020-02-23T17:39:25 |
353395
|
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|
Stack Exchange
|
Holomorphic map proper after shrinking (Kollar's Lecture on Resolution of Singularities)
I'm reading Janos Kolloar's Lecture on Resolution of Singularities and have some problems to understand a detail in the proof of Thm. 1.5 on page 10:
Thm 1.5 (Riemann) Let $F(x,y)$ be an irreducible complex polynomial and $C:=V(F(x,y)) \subset \mathbb{C}^2$ the corresponding compex curve. Then there is a $1$-dimensional complex manifold $\bar{C}$ and a proper holomorphic map $\sigma: \bar{C} \to C$ which is a biholomorphism except at finitely many points.
Proof. Since $F$ irreducible and $\partial F / \partial y$ have only finitely many common points $\Sigma \subset C$. By implicit function thoerem, the first coordinate projection $\pi: C \to \mathbb{C}$ is a local anylytic biholomorphisc on $C \backslash \Sigma$.
We start by constructing a resolution for a small neighborhood of a point $p \in \Sigma$. For notational convenience assume $p=0$, the orgin. Let $B_{\epsilon} \subset \mathbb{C}^2$ denote the ball of radius $\epsilon$ around the origin. Chossing $\epsilon$ small enough, we may assume that $C \cap \{y=0\} \cap B_{\epsilon}= \{0\}$.
Next, by choosing $\eta$ small enough, we can also assume that the restricted map
$$\pi: C \cap B_{\epsilon} \cap \pi^{-1}(\Delta_{\eta}) \to \Delta_{\eta}$$
is proper (???) and a local analytic biholomorphism exept at the origin, where $\Delta_{\eta} \subset \mathbb{C}$ is the disc of radius $\eta$ ...
Question: Why shrinking $\eta$ small enough allows to assume that the restriction of $\pi$ to $C \cap B_{\epsilon} \cap \pi^{-1}(\Delta_{\eta})$ becomes proper map (from topological viewpoint)?
In topology $\pi$ is proper if for every compact $K$ set the preimage $\pi^{-1}(K)$ is also compact.
Some comments: I assume that by $B_{\epsilon}$ and $\Delta_{\eta}$ the author means the open ball resp. disc since in case of closed the subset $C \cap B_{\epsilon}$ would already be compact and it would not be necessary to shrink $\eta$ in "appropriate" way. I have already asked the same question in MSE without obtaining a precise answer that solves the problem.
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2025-03-21T14:48:29.929571
| 2020-02-23T19:23:33 |
353398
|
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|
Stack Exchange
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Lower bound for $[ H : H \cap xHx^{-1} ]$
Let $H$ be a subgroup of a finite group $G$, and let $N = N_G(H)$ be the normalizer of $H$ in $G$.
For $x \in G$ is there a lower bound for $[ H : H \cap xHx^{-1} ]$? If $x \in N$ this index is 1, of course. If $x \notin N$ do we have $[N : H] \leq [H : H \cap xHx^{-1} ]$? What if $N/H$ is cyclic?
So if $H^$ is a center-free group and $H$ is the "diagonal subgroup" of the wreath product of $H^$ and $C_p$ we have $[N : H] = p$ and we can find a group element $x$ such that $H \cap xHx^{-1} = 1$. I see this. In this case, I think, $N = H \times C_p$.
What if $N$ is not a trivial extension of $H$ and $N/H$?
You can make $[N:H]$ as big as you want: start with an arbitrary group $E$ and non-normal subgroup $H$ (e.g. $E=C_p\rtimes C_2$) dihedral of order twice a prime $p$, and $H=C_2$; for every $x\not\in H$ one has $[H:H\cap xHx^{-1}]=2$), and take $G$ to be the direct product of $E$ and any other group $U$, cyclic if you want. Then in $G$, the normaliser of $H$ will contain $U$, so $[N:H] \geq |U|$. If in the above example you take $U$ to be cyclic of order gazillion, then $N=UH$, and $N/H$ will be cyclic of order gazillion.
Edit to address the modification in the comment: merely demanding that the extension of $N/H$ by $H$ is non-split is not enough to eliminate "silly" example as above: take a group $X$ whose automorphism group contains $S_3$, the symmetric group of order $6$, and such that $C_3\leq S_3$ acts fixed point freely on $X$ (e.g. $X=C_2\times C_2$ will do), and take the semidirect product $E=X\rtimes S_3$. Let $H$ be the subgroup of order $3$ in $S_3$. Then for every $x\in E$ that does not normalise $H$ one has $[H:xHx^{-1}]=3$. Now take the direct product $G$ of $E$ and any cyclic group $U$ of odd order. Then the normaliser of $H$ in $G$ is generated by $S_3$ and $U$. Since $U$ is cyclic of odd order, $N/H\cong C_2\times U$ is cyclic, but now the $C_2$ acts non-trivially on $H$, so the extension is not split.
Although in a comment, @AJB clarified (although it is not clear whether it is an ad hoc correction) that they hoped for a case where $N$ was not a split extension of $H$ by $N/H$.
|
2025-03-21T14:48:29.929769
| 2020-02-23T20:57:50 |
353403
|
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|
Stack Exchange
|
Do we know how to determine the $2^{2020}$ decimal of $\sqrt{2}$?
In the case of $\dfrac{1}{7^{800}}$ it's easy, to find the $2^{2020}$ decimal, but what about the simplest of the irrational numbers.
Question: Do we know how to determine the $2^{2020}$ decimal of $\sqrt{2}$?
the algorithm is known --- you may have to wait a long time for it to finish.
@Carlo Beenakker, I think this question requires to mobilize a strategy different from that which you propose in your link.
@Qfwfq Newton-Rapson won't finish before the Sun explodes and won't have enough particles in the visible universe for the required memory storage, but otherwise it is fine. A legitimate answer should produce the required digit without computing all the preceding ones and that may be quite tricky since there is no obvious pattern in the digit sequence. I would say the answer is "no" but there is no proof of that either.
according to arXiv:0912.0303 no digit-extraction formula is known for $\sqrt 2$.
@Carlo : well it's the good forum, research level
This problem has nothing to do with arithmetic geometry. Arithmetic geometry is algebraic geometry over finite fields, local fields, global fields etc.
What does $7^{800}$ have to do with this?
It was to give the example of a rational whose period is greater than $2^{2020}$
@Dattier Yeah, but if it were slightly less than $2^{2020}$ (i.e., if the corresponding decimal digit weren't obviously $0$), it would make another interesting question...
Not voting to reopen because the question seems hopeless, but, of course, the suggestion to send it to MSE (implied by the formal closing reason) is ridiculous.
While there is no known algorithm essentially faster than Newton iteration, one can significantly improve on the “not enough particles in the visible universe” front: the decimal expansion of $\sqrt2$ is computable in logarithmic space, which would mean here something on the order of a few KB. (Better yet, it is computable in logarithmic time on a “threshold Turing machine”, if anyone figures how to build one in the real world.) Timewise, this would be much slower than Newton iteration.
@EmilJeřábek3.0, could you please elaborate? Where is the space saving for storing the digits coming from?
@IgorKhavkine It’s a completely different algorithm. Basically, you compute the square root by partially summing a power series, but the hard part is to compute the relevant iterated additions and multiplications without storing all the terms (recomputing some of it when needed). The basic idea of the iterated multiplication algorithm is to do it modulo many small primes and then reconstruct it from the Chinese remainder representation, but there is more to it than that. See Hesse, Allender, and Barrington https://doi.org/10.1016/S0022-0000(02)00025-9 .
According to Wolfram Alpha there might be chance at base two: https://www.wolframalpha.com/input/?i=series+sqrt%28x%29+at+x%3D2
@fedja Why is the $2^{2020}$th decimal of $7^{-800}$ obviously $0$?
$$d=E(\dfrac{10^{2^{2020}}}{7^{800}}) \mod 10=\left( \dfrac{10^{2^{2020}}-(10^{2^{2020}}\mod 7^{800})}{7^{800}} \right) \mod 10=\dfrac{-(10^{2^{2020}}\mod 7^{800})}{7^{800}} \mod 10$$ and I find $7$.
@EmilJeřábek3.0 Because my brain was totally dead in the evening :-) That happens sometimes. No, it is not. Just ignore that remark...
@fedja No worries. You know, I first asked you about this in a comment, then later when rereading it I realized that it is indeed obviously 0 and deleted my silly comment, and then later I realized that my earlier obvious reason was off by an exponential and reposted the comment again.
Obviously the answer is the digit $3$.
@KonstantinosGaitanas : why?
When I was a student in London I used to attend some lectures by the (great and loveable) C.A. Rogers. Once - having tea and chat with him - someone asked which question he would most like to have answered. His almost immediate response was: "I'd love to know the decimal expansion of the square root of 2"
I have rolled back the edits to this old question, which seem to be purely cosmetic and done for promotion. Yes the question is appropriate for MO, but that doesn't mean you need to add some kind of "RL" label
If you are claiming that you can answer this question, then you can either add an answer below, or write up your work as a preprint in the normal academic way, or write it in a blog post and leave a link. Saying "I can answer this, can you?" is not a constructive use of this site, nor is it collegial.
Well, I would wait 500 years for the math community to find a complex 600-page solution, like for Fermat.
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2025-03-21T14:48:29.930135
| 2020-02-23T21:02:47 |
353404
|
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|
Stack Exchange
|
Hessian matrix and its positiveness
Assume that $\Sigma^2$ is a closed surface in $\mathbb{R}^3$ defined by the equation $\rho(x)=1$, where $\rho$ is some smooth function so that $\nabla \rho\neq 0$. Let $A=H(\rho)$ be the Hessian matrix of $\rho$. My question arises, whether $\left<Ax,x\right>>0$ for $x\neq 0$ implies that the domain $\Omega$ bounded by $\Sigma^2$ is convex?
Convex functions have convex level sets
Yes. your condition implies that the surface is locally convex, and the fact that locally convex implies globally convex is a theorem of Tietze.
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2025-03-21T14:48:29.930205
| 2020-02-23T21:28:17 |
353406
|
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|
Stack Exchange
|
How many flips of a fair coin are needed to get at least one run of at least $k$ consecutive heads with probability $\ge1/2$?
The following question was asked today:
How many flips $n$ of a fair coin are needed to get at least one run of at least $k$ consecutive heads with probability $P_{k,n}\ge1/2$?
The question was negatively received and deleted by the OP. However, the answer to this question is not quite trivial and has been a subject of several studies. This answer will be given below.
According to Mathworld (see also Wikipedia),
$$P_{k,n}=1-\frac{F_{k,n}}{2^n},\tag{1}$$
where, for each natural $k$, $(F_{k,n})_{n=1}^\infty$ is the sequence of $k$-step Fibonacci numbers, defined recursively by the formula
$$F_{k,n}=\sum_{i=1}^k F_{k,n-i}\tag{2}$$
with the initial conditions $F_{k,n}=0$ for $n\le0$ and $F_{k,1}=F_{k,2}=1$.
It is known (see e.g. formula (4)) that for $k\ge2$
$$F_{k,n}=R\Big(\frac{(r_k-1) r_k^{n-1}}{(k+1) r_k-2 k}\Big),\tag{3}$$
where $R(x)$ is the nearest integer to a real number $x$, and $r_k$ is the only root $r$ of the equation $r^k(2-r)=1$ in the interval $(1,2)$.
It follows that for each fixed natural $k\ge2$ we have $F_{k,n}=r_k^{(1+o(1))n}=o(2^n)$ as $n\to\infty$. In view of (1), this implies that the smallest $n$ in question exists and equals
$$n_k:=\min\{n\colon F_{k,n}\le\tfrac12\,2^n\}.$$
The values of $n_2,\dots,n_{10}$, found by Mathematica -- in about 0.1 sec using (2) (storing all the previously found values of $F_{k,n}$), and in about 6 sec using (3) -- are $4, 10, 22, 44, 89, 178, 356, 711, 1421$. Also, obviously, $n_1=1$.
Nice answer: Related: Enumerating binary strings without r-runs of ones, by MA Nyblom.Available via Google scholar profile at https://scholar.google.com.au/citations?user=FxpK36cAAAAJ&hl=en
@kodlu : Thank you for your comment and the reference to Nyblom's paper; I actually had seen it.
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2025-03-21T14:48:29.930345
| 2020-02-23T21:33:26 |
353408
|
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|
Stack Exchange
|
Embedding of a group into a simple group in which every element is a commutator
It is known that any group $G$ can be embedded into a simple group $S$, see, e.g., the discussion at Can any group be embedded in a simple group?
My question is whether one can get an embedding such that the ambient group $S$ is of commutator width $1$ (i.e., each element of $S$ can be represented as a single commutator).
Yes, every infinite group $G$ naturally embeds into the quotient of the symmetric group $S(G)$ by its maximal normal proper subgroup (the subgroup of permutations of support $<|G|$) which is a simple group (Baer) and in which every element is a commutator (because it holds in $S(G)$, in turn because it holds in $S(\omega)$, Ore Proc AMS 1951).
Note: the above embeds a group of cardinal $\alpha$ into a simple group of cardinal $2^\alpha$. Nevertheless, whenever $G\le H$ with $G$ infinite and $H$ simple, then $G$ is contained in a simple subgroup $L$ of $H$ of cardinal $|G|$ (by an easy argument: add finitely many conjugators for each pair of nontrivial elements and iterate). If $H$ has commutator width 1, do the above at odd times and at even times add for every element 2 elements of which they're commutator. In this way, get $L$ in addition of commutator width 1.
Note: the fact that every group embeds into a simple group of the same cardinal is asserted (with hints of proof based on the same principle) in: Locally finite groups by Kegel and Wehrfritz, 1973, Remarks p.115. This follows a detailed proof that in an infinite simple group, every countable subset is contained in an infinite countable simple subgroup (Theorem 4.4 therein). In the same remark they also mention that this is a part of general result of universal algebra.
@YCor: All the comments are greatly appreciated (sorry for unforgivably long delay). Interestingly, similar phenomena take place for associative algebras and Lie algebras (L.A. Bokut', 1960-70's).
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2025-03-21T14:48:29.930508
| 2020-02-23T22:38:34 |
353411
|
{
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"Claudio Rea",
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|
Stack Exchange
|
Discrete superharmonicity
The value at $(n,m)$ of the “Discrete Laplace operator” (see wikipedia) of a function $f$ in $\Bbb Z\times \Bbb Z$ is $\Delta f(n,m)= \frac{1}{4}( f(n+1,m)+f(n,m+1)+f(n-1,m)+f(n,m-1))-f(n,m)$:
the difference between the average of $f$ at the $4$ next points and the value of f at $(n,m)$.
If $g\in C^2(\Bbb R^2)$ we can set
$f_\varepsilon (n,m)=g(x+n\varepsilon,y+n\varepsilon)$, then when $\varepsilon\downarrow 0$, $ f_\varepsilon\to g$ and $\Delta f_\varepsilon\to\Delta g$ the usual Laplacian. Usual harmonic ($\Delta g=0$) and subharmonic
($\Delta g\geq 0$) functions in $g\in C^2(\Bbb R^2)$ when upper-bounded are constant.
Question 1 Upper-bounded discrete harmonic functions $f$ in $\Bbb Z\times \Bbb Z$ are constant? (I have done that)
Question 2 Is that still true for discrete sub-harmonics? (I could not do that but might be in some paper).
Judgments about the banality of the questions are welcome if preceded by their solutions or references.
For the benefit of other readers: I would like to suggest that those who have voted to close have mistaken this question for an exercise, when the text indicates that it is actually a genuine question arising from an easier version that the OP could solve
I will elaborate a more pompous version. This is a difficult problem in discrete harmonic analysis. It might be treated in some paper. Even the harmonic case (which I solved) is absolutely nontrivial. People should solve problems before snubbing them.
@ClaudioRea I voted to close your question; I voted to reopen it. Closing a question doesn't mean snubbing it. It's often done on the belief that the question is an exercise, according to a number of clues (a sketch of proof in the mind, also "I can't do it" in the title which is often evidence for poor level questions, etc.). Such cooperative closing maintains the quality of the site. It happens to accidentally close questions which shouldn't be. The best reaction one can do in such cases (and you did) is to edit the question with additional information/context or improve it.
Regarding problem 1, Deny, Sur l’équation de convolution $\mu = \mu * \sigma$ is a classical reference, I think.
@SamHopkins: Isn't this paper about signed super-harmonic functions?
Problem 2 seems to follow by general theory, although I did not verify the assumptions for the Riesz decomposition theorem. This theorem asserts that every positive super-harmonic function decomposes into the sum of a positive harmonic function and a potential. We already know that harmonic functions are zero everywhere, and in $\mathbb{Z}^2$ the potential is either everywhere zero or everywhere infinite.
If somebody is interested in, I can show my hand-made solution of problem 1
I think there is a discrepancy between the question in the title and the in the body of the question: one is about superharmonic functions (nonnegative, I guess), the other one is about bounded subharmonic functions. Anyway, in both cases one can use the random walk: if you apply your function to the random walk then you get a nonnegative supermartingale (or bounded submartingale); by martingale convergence theorem it converges. But the random walk is recurrent, so the function has to be constant.
I changed super-harmonicity into sub-harmonicity just for having a reference in wiki. I really believe that the reader understands that the only difference is the sign. Frankly this seems to me a very minor discrepancy. Mateusz's arguments are much beyond my culture, I don't doubt that they are very solid. My poor answer to question 1, that I wouldn't dare to publish on M.O., is really down to earth but has the advantage being self-contained and tactile, accessible to everybody. That's probably the reason why I could not extend it to question 2.
|
2025-03-21T14:48:29.930916
| 2020-02-23T23:10:23 |
353413
|
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|
Stack Exchange
|
Exact sequence for the sheaf of principal parts
Let X be a smooth projective curve over a basis S of characteristic $ p$. Denote by $\mathscr P^1_{X/S}$ the sheaf of principal parts of degree $≤ 1$, namely the structural sheaf of the second order infinitesimal neighborhood $∆^{(2)}$ of the diagonal in $X ×_S X$ (EGA IV, sect $16$). The left $\mathscr O_X$-module structure is defined by the first projection and the right one by the second projection.Let 1 be the global section of $\mathscr P^1_{X/S}$ defined by the constant function with value 1 on $∆^{(2)}$ . The
restriction to the diagonal gives the exact sequence
$0 → ω_{X /S} → \mathscr P^1_{X/S} → \mathscr O_X → 0$ of left $\mathscr O_X$-modules.
I am having hard time understanding this paragraph, namely:
$ω_{X /S}$ is supposed to be the canonical line bundle of $X$. I found many references on this notion, for instance here and here but which applies in this context?
From definitions I found that $\mathscr P^1_{X/S} = ∆^* \mathscr O_{X \times_S X}/\mathscr I^2$ for some sheaf of $\mathscr O_X-$ modules $\mathscr I$. But I don't understand the sentence with first and second projections
Finally I don't how we get the exact sequence.
Thank you for your help.
If you don’t like the EGA reference, there’s a very nice explanation of this concept in the book 3264 and All That by Eisenbud and Harris in the chapter on contact problems.
Thank you I will check this book
Actually I didn't find any reference to these questions in EGA IV
See Section 3.2 of "Parabolic connections and stack of roots".
|
2025-03-21T14:48:29.931066
| 2020-02-25T21:24:00 |
353559
|
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|
Stack Exchange
|
Regular triangulation of hypercube
I have started studying regular subdivisions of the $n$-cube, and came across the following post: Regularity of Delaunay triangulation of a hypercube.
My question is whether the "standard triangulation" of the $n$-cube described in the answer of that post is always regular. I know that the answer is yes in two and three dimensions, and, intuitively, I would guess that it is also true for arbitrary $n$. I'd be grateful for any help.
(A subdivision is regular if it can be obtained by lifting the vertices to $\mathbb{R}^{n+1}$, and projecting the lower faces of the convex hull of the lifted vertices to $\mathbb{R}^n$.)
More generally the canonical triangulation of the order polytope of any poset (with simplices corresponding to linear extensions) is regular. But this question seems a little too low-level for this site.
See pp. 439-440 of https://www.csun.edu/~ctoth/Handbook/chap16.pdf.
@SamHopkins Thank you! I‘m new to this topic, and wasn‘t aware that it’s that simple.
|
2025-03-21T14:48:29.931175
| 2020-02-25T21:24:43 |
353560
|
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|
Stack Exchange
|
Absolutely continuous invariant measures for a minimal flow
Fix $(X, g)$ to be some compact Riemannian manifold and $V \in \Gamma(TX)$ a smooth, non-vanishing vector field. Suppose the flow is minimal, i.e. every orbit is dense in $X$, and volume-preserving.
Suppose there exists a measure $\sigma$ that is both invariant under the flow of $V$ and absolutely continuous with respect to the Riemannian volume measure. In other words, it is given by integration against $f \text{dvol}_g$ for $f \in L^1(X)$.
If $f$ is not merely $L^1$ but continuous, then our topological assumptions imply immediately that $f$ is constant, so actually any invariant measures are a constant multiple of the volume measure (the minimality precludes the existence of any singular invariant measures).
Can we construct an example such that $f$ is not constant almost everywhere?
I don't think the minimality precludes the existence of singular invariant measures. You have the celebrated Furstenber's example, in Furstenberg's paper "Strict erogidicity and transformations of the torus", which is a minimal smooth area-preserving diffeomorphism in the 2-torus and is not uniquely ergodic. In that example every ergodic measure is singular respect to Lebesgue.
Ah yes, you're right. The singular invariant measures could be supported on a Lebesgue measure zero, yet dense subset. I did read a bit about the Furstenberg example, but it's explicitly noted that the transformation here is not isotopic to the identity - in particular it can't arise from a flow like in my question.
Furstenberg's example is indeed isotopic to the identity. However, that is not the point. The suspension flow of any minimal diffeomorphism produces a minimal flow and if the diffeomorphism preserves a smooth volume form (in this case an area form), the suspended flow preserves one as well. So, one can produce such a flow from Furstenberg's example.
Ah, you're right. Thanks!
It seems to me that the answer to your question is "yes", and in fact,
Furstenberg's example can be used to construct such an example.
The idea of Furstenberg's construction in the paper Strict Ergodicity and Transformation of the Torus consists in finding an
irrational number $\alpha$ and a $\phi\in
C^\infty(\mathbb{T}^1,\mathbb{R})$ with $\int\phi(x) d\mathrm{Leb}(x) = 0$ such that there is a
measurable $L^1$ function $u\colon \mathbb{T}\to\mathbb{R}$ satisfying
$\phi(x) = u(x+\alpha) - u(x)$ for Lebesgue a.e. $x\in\mathbb{T}$, but
there is no continuous function $u$ satisfying this property.
Then, the diffeomorphism $g : \mathbb{T}^2 \ni (x,y) \mapsto
(x+\alpha,y+\phi(x))$ is minimal, preserves the Lebesgue measure of
$\mathbb{T}^2$, but this is not an ergodic measure. In fact, the map
$h : (x,y)\mapsto (x,y+u(x))$ leaves invariant the Lebesgue measure of
$\mathbb{T}^2$ and $g\circ h = h \circ k$, where $k : (x,y)\mapsto
(x+\alpha, y)$; and Lebesgue is clearly not ergodic for $k$.
However, there are infinitely many $k$-invariant measures which are
absolutely continuous with respect to Lebesgue. To see this, one can
consider for instance a non trivial interval $I\subset\mathbb{T}$
(i.e. $I$ and its complement contain more than one point), and
define the measure $\mu:= \mathrm{Leb} \otimes
\frac{1}{\mathrm{Leb}(I)}\mathrm{Leb}\big|_I$. Then, since $h$
preserves Lebesgue, we conclude that $h_*\mu$ is $g$ invariant, is
absolutely continuous with respect to Lebesgue and is different from
Lebesgue.
|
2025-03-21T14:48:29.931414
| 2020-02-25T22:06:09 |
353561
|
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|
Stack Exchange
|
Representation theory of 3-step nilpotent finite groups
I am interested in understanding the representation theory of certain finite nilpotent groups (over the complex numbers). The groups $G$ of interest have the following properties:
1) G is $3$-step nilpotent finite group
2) G is a $2$-group
3) G is an extension of a 2-step nilpotent group H, by $\mathbb{Z} / 2 \mathbb{Z}$.
I know that the irreducible representations of a finite nilpotent group are monomial, and there is a uniform construction of these representations in the case of $2$-step nilpotent groups. For a $2$-step nilpotent group $H$, each irreducible representation $\rho$ of $H$ of dimension greater than $1$ is induced from a character of a fixed maximal abelian subgroup $S$ of $H$. In fact $\rho$ is determined by its central character.
Question:
What can be said about the representation theory of $G$? Can we identify a minimal set of the subgroups that can be used to construct each irreducible representation as a monomial representation? (For example, for a $2$-step nilpotent group we need a maximal abelian subgroup and the group itself.)
I would appreciate any thoughts on this question, any references on 1-3 above. That is, any references on the representation theory of $3$-step nilpotent groups, $2$-groups, or two-fold extensions of $2$-step nilpotent group, or any combination thereof.
Sincerely,
Moshe
An extension of $A$ by $B$ has two possible meanings (is $A$ the kernel or the quotient? unfortunately both conventions are widely used).
Apologies. An extension of $A$ by $B$ means that $A$ is the quotient.
|
2025-03-21T14:48:29.931549
| 2020-02-25T23:34:21 |
353564
|
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|
Stack Exchange
|
Birkhoff's completeness theorem put into practice
Birkhoff's completeness theorem (see here, Theorem 14.19) states that an equation which is true in all models of an algebraic theory can be proven in equational logic.
Question. Does the proof of Birkhoff's completeness theorem actually produce for each specific equation a proof in equational logic? If yes, can you please demonstrate this with an instructive example?
Actually, I suspect that the answer is "No", but I am not entirely sure.
Let us look at the following well-known statement: If $R$ is a ring in which every element $r \in R$ satisfies $r^2=r$ (i.e. $R$ is boolean), then $R$ is commutative. There is the following overkill proof: $R$ is reduced, hence a subdirect product of domains $R_i$. Since the maps $R \to R_i$ are surjective, each domain $R_i$ satisfies the same equation and then must be isomorphic to $\mathbb{F}_2$. In particular, $R_i$ is commutative. Since $R \to \prod_{i \in I} R_i$ is injective, it follows that $R$ is commutative. So by Birkhoff's theorem, there must be some equational proof of this. My question is not how an equational proof looks like - this is just a basic algebra exercise. I would like to see how (if possible) it can be extracted from the overkill proof.
I think the proof of Birkhoff's theorem in this special case works as follows: Consider the free ring on two generators $\mathbb{Z}\langle X,Y\rangle$ and take the quotient with respect to the relations $r^2=r$ for all elements $r$. This is the free boolean ring $R$ on two generators. By the overkill proof, $XY=YX$ holds in $R$. This means that $XY=YX$ can be derived from the relations $r^2=r$ in $\mathbb{Z}\langle X,Y\rangle$. But we don't get a derivation, right? How to produce, for example, the following equation?
$$\begin{align*}
XY - YX &= \bigl((X+Y)^2 - (X+Y)\bigr) - \bigl(X^2-X\bigr) - \bigl(Y^2-Y\bigr) \\
&\phantom{=}+ \bigl((YX)^2 - (YX)\bigr) - \bigl((-YX)^2 - (-YX)\bigr),
\end{align*}$$
As far as I can tell, we are not even guaranteed a priori to have a proof which works without the axiom of choice, since this is used in the structure theorem for reduced rings in the overkill proof? (At first this might be confusing since the axiom of choice is surely not allowed in an equational proof, but actually the axiom of choice is used to show the existence of an equational proof.)
I am interested in more complicated applications of Birkhoff's theorem. This here is just an example to get started. You can also choose other examples if they are more instructive.
There should be varieties with nonrecursive (but r.e.) equational theories. Short of doing computations in a finitely generated relatively free algebra, I am not sure how to answer your question when it can be answered at all. Gerhard "Equational Proof Is Just Equations" Paseman, 2020.02.25.
It is an old theoren of Jacobson that if R if a ring satisfying the equation $x^n=x$ is commutative. The prof uses subdirect products and structure of division rings. There are very few examples of $n$ where an equational proof of commutativity is known. Birkhoff’s proof uses choice I believe and doesn’t really tell you how to get equations.
@BenjaminSteinberg Actually, I have an equational proof for infinitely many $n$. :-)
@MarkSapir Interesting. Could you please give a reference for the syntactic proof?
All right, thank you.
@MarkSapir, Jacobson's identity means the ring is von Neumann regular (in fact completely regular as a semigroup). This trivially forces the radical to be zero since every non-zero left ideal of a von Neumann regular ring has a nonzero idempotent and the radical cannot have a nonzero idempotent since they are not left quasiregular.
Let me try to restate the question.
I consider an identity to be a pair, written $(s,t)$ or $s\approx t$. I also consider a set of identities to be a set of pairs.
Birkhoff's Theorem compares three things, namely
(1) $\Sigma\models s\approx t$,
(2) The pair $(s,t)$ belongs to the fully invariant congruence $\Theta^{T(X)}(\Sigma)$ on the term algebra $T(X)$, and
(3) $\Sigma\vdash s\approx t$.
I think the question is asking whether it is possible to extract from the proof of Birkhoff's Theorem and from a specific instance of $\Sigma\models s\approx t$ a proof witnessing that $\Sigma \vdash s\approx t$.
The proof of Birkhoff's theorem does explain why (1) $\Leftrightarrow$ (2). Also, if we are told HOW the pair $(s,t)$ is generated as a member of $\Theta^{T(X)}(\Sigma)$, that can be translated into a $\Sigma$-proof of $s\approx t$. What is missing is: given the knowledge that
$(s,t)\in\Theta^{T(X)}(\Sigma)$, do we know how the pair $(s,t)$ got into $\Theta^{T(X)}(\Sigma)$?
The proof of Birkhoff's Theorem does not require this missing ingredient, it only requires that if $(s,t)$ is in $\Theta^{T(X)}(\Sigma)$, then it was generated in some way. So the answer to the question in the form it was asked (can we extract a $\Sigma$-proof of $s\approx t$ from the proof of Birkhoff's Theorem?) has to be No.
You could try asking a slightly different question: Suppose, given finite $\Sigma$, that we are told that $\Sigma \models s\approx t$. By B's Theorem we know there must exist a $\Sigma$-proof of $s\approx t$. Question: can we estimate an upper bound on the complexity of such a proof?
An affirmative answer would imply that any finitely axiomatizable variety has a decidable equational theory. But we know this to be false, since in the 1940's Tarski found a finitely axiomatizable variety of relation algebras with an undecidable equational theory. Many other examples are known now.
Thank you! So my guess was correct.
The proof of Birkhoff's theorem, much like the proof of the completeness theorem for predicate logic, will not give a very interesting algorithm for computing a proof for a statement known to be valid: it will essentially consist of a naive proof search which, for a valid statement, will terminate. But as Keith Kearnes wrote, there cannot be a computable bound for the size of a proof in the size of the statement.
However, note that Birkhoff's theorem uses only the truth of the statement under consideration. It does not use any information contained in any of the known proofs of the statement. In proof theory there are techniques for using this information, in particular Gentzen's cut-elimination theorem. It provides a proof transformation which, in some sense, allows to remove notions for a proof which do not appear in the theorem it shows. Famous applications of the cut-elimination theorem to mathematical proofs are Girard's transformation of the Fürstenberg-Weiss proof of van der Waerden's theorem into the original proof [1, annex 4.A] and Luckhardt analysis of Roth's theorem about approximations of irrational numbers [2].
A strongly related line of work is proof mining [3] which focuses on the extraction of functionals and bounds from proofs.
Instead of applying Birkhoff's theorem, it may be more fruitful to apply such proof-theoretic techniques since they also allow to leverage the proof of the statement under consideration.
[1] J.Y. Girard: Proof Theory and Logical Complexity. In Studies in Proof
Theory, Bibliopolis, Napoli, 1987.
[2] Horst Luckhardt. Herbrand-Analysen zweier Beweise des Satzes von
Roth: Polynomiale Anzahlschranken. Journal of Symbolic Logic,
54(1):234–263, 1989.
[3] Ulrich Kohlenbach. Applied Proof Theory: Proof Interpretations and
their Use in Mathematics. Springer, 2008.
|
2025-03-21T14:48:29.932066
| 2020-02-26T01:21:42 |
353568
|
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|
Stack Exchange
|
A ‘canonical’ bounded lattice with proper de Morgan negation?
Call a lattice negation $\neg$ proper de Morgan negation iff it satisfies the following conditions.
$\neg\neg a=a$.
$\neg(a\vee b)=\neg a\wedge\neg b$ and $\neg(a\wedge b)=\neg a\vee\neg b$.
$a\leqslant b\Leftrightarrow\neg b\leqslant\neg a$.
If $a\vee\neg a=\top$, then $a\in\{\top,\bot\}$.
Consider now the following lattice and call it $\mathbf{M}^{\omega\omega^*}_{\omega}$. In this lattice, there are $\omega$ chains each of which is order-isomorphic to $\omega+\omega^*$ with $\omega^*=\langle\ldots,n,\ldots,0\rangle$.
It seems that if $\neg\mathbf{i_j}=\mathbf{i_{-j}}$, the negation is a proper de Morgan one.
The question now is whether the following statement holds.
Let $\phi$ and $\chi$ be propositional formulas over $\{\neg,\wedge,\vee\}$. Define $v_{\mathfrak{L}}$ (a valuation) as a function from the set of propositional variables to the underlying set of lattice $\mathfrak{L}$ and extend it to complex formulas in a straightforward way.
Say that $\phi$ entails $\chi$ in $\mathfrak{L}$ ($\phi\vDash\chi$) iff for any $v_{\mathfrak{L}}$: $v_{\mathfrak{L}}(\phi)\leqslant v_{\mathfrak{L}}(\chi)$.
Then for any bounded lattice $\mathfrak{L}$ with a proper de Morgan negation the following holds $$\phi\nvDash_{\mathfrak{L}}\chi\Rightarrow\phi\nvDash_{\mathbf{M}^{\omega\omega^*}_{\omega}}\chi$$
At first, it seemed correct to me and I (unsuccessfully) tried to prove it as follows.
Consider a valuation $v$ in $\mathfrak{L}$ that refutes $\phi\vDash_{\mathfrak{L}}\chi$ and construct a valuation $v'$ in $\mathbf{M}^{\omega\omega^*}_{\omega}$ as follows.
If $v(p)=\top,\bot$, then $v'(p)=\top,\bot$, respectively.
If $v(p\vee q)=\top$, then $v'(p)$ and $v'(q)$ are incomparable w.r.t. order on $\mathbf{M}^{\omega\omega^*}_{\omega}$.
If $v(p)\leqslant v(q)$, then $v'(p)\leqslant v'(q)$.
And here begins the problem with the last case: $v(p)$ and $v(q)$ are incomparable in $\mathfrak{L}$ but $v(p\vee q)\neq\top$. It is impossible to find a valuation $v'$ on $\mathbf{M}^{\omega\omega^*}_{\omega}$ satisfying these conditions.
However, I don't understand whether it follows from here that the statement of the theorem fails.
As an aside, do I understand it correctly that in case the theorem is incorrect, it would likewise fail for the same lattice but with chains order-isomorphic to $\omega+1+\omega^*$, i.e. $\langle 1,\ldots,n,\ldots,0,\ldots,-n,\ldots,-1\rangle$?
Reposted from MSE: https://math.stackexchange.com/questions/3552688/a-canonical-bounded-lattice-with-proper-de-morgan-negation
Let $\phi=(p\vee(q\wedge r))\wedge (r\vee(p\wedge q))$ and let $\chi = (p\wedge (r\vee (p\wedge q)))\vee (r\wedge (p\vee (q\wedge r)))$. For your De Morgan lattice ${\bf M}$ we have $\phi\models_{\bf M} \chi$, but we have $\phi\not\models_{\bf L} \chi$ for the De Morgan lattice:
The complementation is defined by the self-duality that fixes $p$ and $r$.
This example is more complicated than Adam's example, but it has the property that it is purely about the underlying lattice.
The justification is this: Let ${\bf L}^*$ be ${\bf L}$ minus its top and bottom. ${\bf L}^*$ is the splitting lattice for the $p$-modular law, which is $\phi\leq \chi$. Any lattice either satisfies the $p$-modular law, or has a copy of ${\bf L}^*$ as a sublattice. Your lattice ${\bf M}$ doesn't have a sublattice isomorphic to ${\bf L}^*$, so it satisfies the $p$-modular law. On the other hand, ${\bf L}$ does not satisfy the $p$-modular law, as you can see by assigning to the variables $p, q, r$ the values indicated in the figure.
Notice that if ${\mathbf L}$ is any bounded lattice, then the ordinal sum ${\mathbf 1}+{\mathbf L}+{\mathbf L}^{\partial}+{\mathbf 1}$ has a proper De Morgan complementation satisfying all four bullet points of the problem. Any canonical example ${\mathbf M}$ would have to fail every lattice identity that failed this ordinal sum, hence every lattice identity that failed in ${\mathbf L}$. Since ${\mathbf L}$ is arbitrary, it follows that the underlying lattice of any canonical example would have to generate the variety of all lattices. Since your lattice is uniformly locally finite, it cannot generate the variety of all lattices.
The statement that you are trying to prove appears to be false. In your algebra the inequality $x \wedge (y \vee \neg y) \leq (x \wedge y) \vee (x \wedge \neg y)$ holds, but it does not hold in every lattice with a De Morgan negation. For a counter-example, consider the five-element diamond $M_3$ equipped with a De Morgan negation which has exactly one fixpoint.
Your condition that $a \vee \neg a = \top$ implies $a = \top$ or $a = \bot$ in fact has no effect on the consequence relation. This is because each lattice with a De Morgan negation can be embedded into a lattice with (in your terminology) a "proper" De Morgan negation just by adding a new top and bottom element to the lattice. The above counter-example can therefore be extended to a "proper" one.
(By the way, the proper spelling of the name is indeed "De Morgan" rather than "de Morgan".)
Thanks! Do I get it correctly that $x\wedge(y\vee\neg y)=(x\wedge y)\vee(x\wedge\neg y)$ holds if and only if the lattice does not have $x$, $y$, and $z$ pairwise incomparable w.r.t. $\leq$ such that $\neg y=z$ and $\neg z=y$?
Not quite: you can find such $x$, $y$, $z$ in a large enough Boolean algebra.
I meant lattices with De Morgan, not Boolean negations, and without distributivity. Or will it also fail?
You can embed a Boolean algebra into a non-distributive lattice with a De Morgan negation by adding some non-distributive lattice (say, $M_3$) with a De Morgan negation to the top and bottom of your lattice, if you want an example which is not distributive.
|
2025-03-21T14:48:29.932432
| 2020-02-26T01:26:13 |
353569
|
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|
Stack Exchange
|
Probability of m crossings of 0 before time 1 of a standard Brownian motion
Let $B$ be a standard Brownian motion. Could anyone show some hints and reference about how to compute the following probability?
Let $N(n) = \sum_{i=1}^n \mathbb{1}_{\{0 \in B[\frac{i-1}{n},\frac{i}{n}] \}}$, Then for any $m <n$, what is
$$
\mathbb{P}\left( N(n) < m \right)?
$$
I have already known that $\mathbb{P}(\exists t\in (a,b): B(t)=0) = \frac{2}{\pi}\arctan\sqrt{\frac{b-a}{a}}$, so $\mathbb{E}\left( N(n) \right) \approx \sqrt{n}$. What I really want to proof is that for any $\epsilon > 0$, $\mathbb{P}\left( N(n) < n^{\frac{1}{2}(1-\epsilon)} \right)$ is summable.
Isn't that essentially the same as the probability that $M_1\le n^{-1/2-\epsilon}$ where $M_t=\max_{0\le s<t}X_t$? If so, the distribution of $M_1$ can be found under running maximum at https://en.wikipedia.org/wiki/Wiener_process>.
@AnthonyQuas Hi, Anthony, I think there is a notation problem in $N(n)$, so we are talking about different things. I have fixed it. Sorry for this notation. I want to talk about the crossings around $x$-axis.
With probability one, the dimension (both Hausdorff and box-counting) of the zero set is $\tfrac{1}{2}$, so with probability one we have $\log N(n,\omega) \sim \log n^{1/2}$. By Fubini's theorem, this implies summability of $\mathbb{P}(N(n,\omega) < n^{(1 - \epsilon) / 2})$. In fact, if I remember correctly, the $\tfrac{1}{2}$-dimensional box-counting measure of the zero set is equal to the local time, so we have $N(n, \omega) \sim L(\omega) n^{1/2}$, where $L(\omega)$ is the local time of $B(t)$ at zero.
I was wrong with the last statement: a correct one seems to be something like $N(n,\omega) \sim L(\omega) \psi^{-1}(1/n)$, where $\psi(h) = h^{-1/2} (\log \log h^{-1})^{1/2}$, at least for the Hausdorff dimension variant; see Theorem 1.2 in Baraka–Mountford, The Exact Hausdorff Measure of the Zero Set of Fractional Brownian Motion, J. Theor. Probab (2011) 24: 271–293.
@MateuszKwaśnicki Hi, Mateusz, thank you for your comment. Could you please explain a little more where you apply the Fubini's theorem? Moreover, It is clear that $N(n, \omega) > L(\omega) n^{1/2(1-\epsilon)}$ a.s., but to get my conclusion, we don't know that $L(\omega)$ is uniformly bounded in [0,1] independent of $\omega$. In fact it may be achieved by compute the probability of the local time no larger than $n^{1/2 \epsilon}$.
@WenboLi: You are right, summability of $\mathbb{P}(N(n) < n^{1/2(1-\epsilon)})$ requires some uniformity with respect to $\omega$. Sorry for that error, I will try to find a workaround.
|
2025-03-21T14:48:29.932635
| 2020-02-26T02:03:54 |
353570
|
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|
Stack Exchange
|
Is there a foundational approach that takes "structure" as primitive?
As per the title, I'd be curious to know if there have been attempts at constructing a foundation of mathematics taking, somehow, purely the notion of "structure" as primitive, maybe via a system of axioms that formalizes how structure can be transported, restricted, extended, encoded, etc.
I'm being intentionally vague about all the above.
I am aware of the existence of structural set theories, such as ETCS which formalizes the notion of the category of sets in the language of category theory, but I'm unsure whether to consider this really a theory "about structure" rather than just about sets seen from an intrinsic non-materialistic viewpoint. Maybe somebody has some arguments in favor of actually seeing it as a "theory of structure".
Same doubts about (homotopy or not) type theories.
Can you expand on how type theory fails to meet your desiderata? I'd argue that taking "structure" as primitive is precisely what type theory does. For instance, in type theory, the natural numbers $\mathbb N$ are defined via a universal property.
@TimCampion I'd expect such a foundation to use a word at least similar to "structure" (as an internal primitive) and to use it in such a way that it reflects our intuition of structures. Are you suggesting type theory satisfies this if we read the word "type" as "structure"? If I read $\mathbb{N} : \text{Set}$ as "the naturals have the structure of a set" there seems to be something missing. How would I talk about "the whole structure of $\mathbb{N}$" inside type theory? (This reminds me that Ocaml uses the keyword struct.)
I expanded on the "types as structure" point of view, and how I view it as related to "structural set theories" (perhaps a misnomer from this perspective), at https://golem.ph.utexas.edu/category/2013/01/from_set_theory_to_type_theory.html . Obviously the word "type" cannot be grammatically simply substituted for "structure", but it has roughly the same meaning.
@MikeShulman Actually... I can't think of an example where I couldn't simply use the word "structure" in place of "type"... Maybe I'd also use "point" instead of "term" if I were doing this... But I'm always a bit confused about what this "pre-theoretical notion of 'structure'" is supposed to be, so maybe I'm missing something.
@MichaelBächtold I don't know where "$\mathbb N: Set$" i coming from. In my understanding of type theory, one would typically have $\mathbb N: Type$, which I think could easily be read as $\mathbb N: \text{Structure}$ -- "the naturals are a point of the structure of structures", i.e. a structure. I'd concede, though, that structuralism is not really among the philosophical perspectives traditionally emphasized by type theorists -- it's not traditionally presented as a theory of structures, so it's surprising (but true, I think) to realize that it could perfectly well be presented this way.
You may also be interested by Steve Awodey's "Structuralism, Invariance, and Univalence" article, at http://www.andrew.cmu.edu/user/awodey/preprints/siu.pdf
|
2025-03-21T14:48:29.933003
| 2020-02-26T02:09:58 |
353571
|
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|
Stack Exchange
|
Two algebraic guises of Alternating Sign Matrices: any connection?
Alternating Sign Matrices (ASMs) have a famous history: they were discovered by Mills, Robbins, and Rumsey, who conjectured a product formula for their enumeration; this product formula was first proved by Zeilberger via a very complicated inductive argument; then Kuperberg gave a much more concise proof using a connection to statistical mechanics, specifically, the so-called "six-vertex model."
The six-vertex model is an exactly solvable model: it can be solved via a (quantum) Yang-Baxter equation. Although certainly this isn't how Baxter thought of it, this Yang-Baxter equation could be thought of as coming from the R-matrix associated to (a particular module for) the quantum group $U_q(\widehat{\mathfrak{sl}}_2)$ (see e.g. here). Thus in at least a tenuous sense ASMs "have to do with" the quantized enveloping algebra of the affine Lie algebra $\widehat{\mathfrak{sl}}_2$.
On the other hand, there is a completely different algebraic guise of the ASMs due to Lascoux and Schützenberger: the ASMs naturally correspond to the elements of the MacNeille completion of the (strong) Bruhat order on the symmetric group. (As far as I know, this algebraic interpretation does not help in any way to count the ASMs.)
Question: is there any direct connection between these two parts of algebra in which ASMs "arise"? Is there any reason to think that the MacNeille completion of Bruhat order on $S_n$ would be related to $U_q(\widehat{\mathfrak{sl}}_2)$?
|
2025-03-21T14:48:29.933142
| 2020-02-26T02:19:32 |
353572
|
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|
Stack Exchange
|
The geometry of lambda calculus?
I stumbled upon "the geometry of quantum computation" --- to quote the abstract:
Determining the quantum circuit complexity of a unitary operation is closely related to the problem of finding minimal length paths in a particular curved geometry
Which is extremely interesting, since one can hope that we can develop techniques to find optimal unitaries that represent an operation. Indeed, the abstract continues:
... We develop many analytic solutions to the geodesic equation, and a set of invariants that completely determine the geodesics. We investigate the problem of finding minimal
geodesics through a desired unitary,
U, and develop a procedure which allows us to deform the
(known) geodesics of a simple and well understood metric to the geodesics of the metric of interest
in quantum computation
My question is whether there exists a similar geometric structure for lambda calculus? Trivially, I can think of "embedding" classical computation inside a quantum circuit by using the usual Toffoli gate trick of producing the inputs and the output to make it reversible.
However, perhaps the geometry of classical computation is more "rigid" than the quantum world, which would be very interesting. I'd like pointers to theories that try to study lambda calculus using differential geometry, in the spirit of the above paper.
|
2025-03-21T14:48:29.933265
| 2020-02-26T05:37:07 |
353580
|
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|
Stack Exchange
|
How geometry changes up to Hermitian inner product on Line bundle (Kodaira embedding)
Riemann metric $g \colon= \Sigma g_{ij} dx_i \otimes dx_j$ on a Kähler manifold $M$ will define the length of a line on $M$, i.e. intrinsic geometry. The line bundle $L$ on $M$ is equipped with a Hermitian inner product $<\,\,,\,\,>$, from which the connection $D$ as well as the associated curvature form $\Omega$ is defined. Often for the Kähler metric $\omega(u,v) \colon= g(Ju,v)$, where $d\omega = 0$, Kodaira embedding requires $\omega = \Omega$.
It seems that the geometry of the Riemann manifold $M$ has something to do with not $L$ only but with the Hermitian inner product $<\,\,,\,\,>$ on $L$. However, algebraic geometry simply defines an ample line bundle $L \colon= L(D)$ on a scheme $X$ as $D$ having positive intersection number with all irreducible curve $C$ on $X$. It seems that this property is intrinsic for $D$, i.e. $L$ itself. That is, $L$ is directly connected with the geometry of $X$.
Q. Why Kodaira embedding requires Hermitian inner product $<\,\,,\,\,>$ on $L$ to equip $L$ the geometry of $M$, i.e., ampleness, whereas algebraic geometry does not?
Your definition of ample is incorrect.
A line bundle $L$ on a complex manifold $X$ is positive if it admits an hermitian metric whose curvature form is a positive (hence, Kähler) form. That form has nothing to do with a given Kähler metric on $X$ as you seem to suggest.
|
2025-03-21T14:48:29.933377
| 2020-02-26T07:00:49 |
353584
|
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|
Stack Exchange
|
Singularity of complete intersection lying on a plane
Let $X$ be a very general hypersurface of degree $\ge 5$ and $Y$ be an irreducible cubic hypersurface in $\mathbb{P}^3$. It is known that $X \cap H$, where $H$ is a hyperplane, can have at most $3$ nodes. My question is: can $X \cap Y$ have more than $3$ nodes lying on a plane $H$ ?
I guess that is not possible. I have the following argument for that.
If possible, let there are at least $4$ nodes of $X \cap Y$ lying on a plane. Choose two of them and consider the line $l$ joining them. Then $l$ is either tangent to $Y$ on lie in $Y$. But a line intersects a cubic curve at at most $3$ points, a contradiction (as $l$ is tangent to $Y$ at two points, $l$ intersects $Y$ at $4$ points counted with multiplicity). Thus $l$ is contained in $Y$. Since $l$ is also tangent $X$ and contained in $H$, $H$ is tangential to $X$ at those points. Choosing another $2$ points and arguing similarly one can see that $H$ is tangent to $X$ at more than $3$ points, a contradiction. Please correct me if I am making any mistake.
|
2025-03-21T14:48:29.933474
| 2020-02-26T09:32:52 |
353589
|
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|
Stack Exchange
|
Definition of reducible lattice
I am reading Raghunathan's book on discrete subgroups of Lie groups.
In particular I am stuck on Corollary 5.19 which gives several equivalent conditions for a lattice in a semisimple Lie group to be reducible.
Recall that a lattice $\Gamma$ in a Lie group $G$ is a discrete subgroup of fintie covolume.
The result I do not understand reads as follows:
Corollary 5.19: Let $G$ be a connected semisimple Lie group without compact factors. Let $H$, $H^{\prime}$ be connected closed proper normal subgroups of $G$ such that $HH^{\prime}=G$, and $H\cap H^{\prime}$ is discrete in $G$. Let $\pi$ (resp. $\pi^{\prime}$) be the natural projection maps of $G\to G/H^{\prime}$ (resp. $G/H$). Let $\Gamma\subset G$ be a lattice then the following are equivalent:
1) $\pi(\Gamma)$ is a discrete subgroup of $G/H^{\prime}$
2)$\pi^{\prime}(\Gamma)$ is a discrete subgroup of $G/H$
3) $\Gamma\cap H$ is a lattice in $H$
4) $\Gamma\cap H^{\prime}$ is a lattice in $H^{\prime}$
There are three more equivalent conditions also but I understand the implications of those as they are spelled out in detail.
What I do not understand are the implications $1) \iff 4)$ and $2) \iff 3)$. What I do understand is that the proofs will be identical.
The book says the following in regards to the proof:
``The implications $1) \iff 4)$ and $2) \iff 3)$ are simply restatements of Theorem 1.13''.
Theorem 1.13 reads:
Theorem 1.13: Let $G$ be a second countable locally compact group and $\Gamma$ a lattice in $G$. Let $H$ be a closed subgroup. Then if $H\cap\Gamma$ is a lattice in $H$, $H\Gamma$ is closed in $G$; equivalently the natural injection
$H/(H\cap \Gamma)\to G/\Gamma$
is proper. Futhermore, if $H$ is normal in $G$ or if $\Gamma$ is uniform (i.e. $G/\Gamma$ is compact), then $H\Gamma$ is closed in $G$ if and only if $H\cap\Gamma$ is a lattice in $H$.
Recall a map is proper if the preimage of compact sets is compact.
I understand Theorem 1.13 and its proof perfectly.
However I cannot see how the mapping $H/H\cap \Gamma\to G/\Gamma$ being proper should imply that the projection $\pi(\Gamma)$ of a lattice is discrete in $G/H^{\prime}$ (and vice versa), similarly for $\pi^{\prime}(\Gamma)$ in $G/H$?
How do compact sets become relevant?
Why should the image of a discrete set stay discrete?
This has been crossposted from stackexchange.
The corollary is false as stated. You need to assume normality of $H,H'$.
@UriBader edited to account for this, thanks
|
2025-03-21T14:48:29.933650
| 2020-02-26T09:36:10 |
353590
|
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|
Stack Exchange
|
Foliation of tangent bundle arising from exponential map
We first mention our motivation:
For $M=\mathbb{R}$ with usual Riemannian metric the exp map $exp:TM\to M$ is in the form$(x,v)\mapsto x+v$
The level sets of this map define a foliation whose leaves are totally geodesic submanifolds with respect to Sasaki metric of $TM$. They intersect the zero section transversaly. The number of intersections with the zero section is the same for all leaves. They are parallel in the sense that for every two points $x,y$ in $M\subset TM$ and every $M$- geodesic curve $\gamma $ joining $x$ to $y$, the tangent space of the leaf passing $x$ is parallel transported along $\gamma$ to thats of $y$.
How can one generalize some or all part of this situation in the case of an arbitrary (compact) Riemannian manifold? Under which conditions on Riemannian manifold, all or some parts of these situations are the case?
I think that the appropriate generalization is that of a tubular neighborhood of a codimension 1 hypersurface in $M$. If the hypersurface $P$ of $M$ is compact, then there exists an open set $U$ in $M$ which contains $P$ and the tangent bundle admits an orthogonal decomposition $$TU=TP \oplus NP$$ where $NP$ is the normal bundle of $P$. An analagous statement to the Gauss lemma shows that this induces a foliation of $U$ whose leaves are hypersurfaces "parallel" to $P$. For a deeper look, I would suggest looking at Lee's new edition of Riemannian manifolds, specifically the discussion of semigeodesic coordinates.
Thank you very much for your answer, your reference to Lee's book and your attention to my question.
|
2025-03-21T14:48:29.933800
| 2020-02-26T12:43:32 |
353599
|
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|
Stack Exchange
|
Lipschitz-continuity of convex polytopes under the Hausdorff metric
Recently, I proved the following Lipschitz-continuity like result for convex polytopes:
Let $A\in\mathbb R^{m\times n}$ and $b,b'\in\mathbb R^m$ be given such that $\{x\,:\,Ax\leq 0\}=\{0\}$ (which is equivalent to boundedness of all induced polytopes) and that $\{x\in\mathbb R^n\,:\,Ax\leq b\},\{x\in\mathbb R^n\,:\,Ax\leq b'\}$ are non-empty. Then
$$
\delta\big(\{x\in\mathbb R^n\,:\,Ax\leq b\},\{x\in\mathbb R^n\,:\,Ax\leq b'\}\big)\leq \Big( \max_{A_0\in\operatorname{GL}(n,\mathbb R),A_0\subset A}\|A_0^{-1}\| \Big)\|b-b'\|_1
$$
where the operator norm $\|\cdot\|$ as well as the Hausdorff metric $\delta$ are taken with respect to $(\mathbb R^n,\|\cdot\|_1)$. Also $A_0\subset A$ is short for "every row of $A_0$ is also a row of $A$" so the above maximum is taken over all invertible submatrices of $A$.
What this intuitively means is that if two polytopes have parallel faces (i.e. they are both described by the same $A$ matrix), but the location of these faces differs ($b\neq b'$), then the distance between the polytopes is upper bounded by the distance between the vectors $b,b'$ times a "geometrical" constant coming from $A$.
Hence the function $b\mapsto\{x\in\mathbb R^n:Ax\leq b\}$ (with suitable domain such that the co-domain equals all non-empty subsets of $\mathbb R^n$) is Lipschitz continuous with a constant determined by $A$.
This came up as a lemma to something only vaguely related which is why I don't have a problem with posting it publicly. Actually if this was a known result then my manuscript could be shortened by 3 pages. Thus my quesiton is:
Is the above result known and, if so, where can in be found in the (convex polytope-)literature?
I would be surprised if nobody has thought about this until now. While I haven't seen this result in the books of Grünbaum and Schrijver or the few papers on convex polytopes I am aware of, this is not the field I usually work in; hence this might very well be known but beyond my mathematical horizon. Thanks in advance for any answer or comment!
It might be helpful if you give a more intuitive, in words, summary of what the continuity result means.
@SamHopkins Thank you for your comment, I think that's a great idea! I just edited my question accordingly.
This is a classic question in the literature on linear programming, since it is related to the stability of the feasible set (and hence the solutions) under perturbation of the parameters.
The classic work in this field is:
A. J. Hoffman, Approximate solutions of systems of linear inequalities, J. Res. Nat. Bur. Standards, 1952.
A more recent work, which essentially states your result as a special case (see Theorem 2.4 and the Lipschitz constant on p.19), is:
W. Li, The sharp Lifshitz constants for feasible and optimal solutions of a perturbed linear program, Linear Algebra and Its Applications, 1993.
This is exactly what I was looking for, thank you very much!
|
2025-03-21T14:48:29.934024
| 2020-02-26T12:51:30 |
353601
|
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|
Stack Exchange
|
Rule to determine rotationally invariant orders of the points of arbitrary 2d splines
I would like to find a rule to determine the order of the points of arbitrary 2d splines, which should be invariant with respect to rotation (as far as possible).
To illustrate the problem, let us imagine a stick figure.
No matter how the arms are positioned and no matter how the stick figure is rotated, we can always decide which hand is the left hand and which hand is the right hand. If we now add the rule that the left hand should always be labeled 1 and the right hand should always be labeled 2, then we have a consistent ordering which is not altered by rotation of the stick figure.
Unfortunately, unlike our stick figure, splines do not have a head, which greatly helps us decide on the order of the hands in case of the stick figure. Therefore, we have to come up with our own reference points.
So the question partly boils down to: What are suitable reference points?
One point that comes to mind, is the middle point of the spline, i.e. the point that cuts the spline in two equally long pieces (comparable to the point of the stick figure that connects the arms).
However, the second point (the head) is more difficult. Any point that would not be rotated along with the spline during a rotation is not suitable. One possibility would be the center of the bounding box of the spline.
I think that this setup would allow a rotationally invariant ordering of the points. However, especially for straight splines, the two reference points are very close to each other, which would make the order of the points very susceptible to slight changes of the curvature of the spline. I.e. the order would be flipped if we bent a straight spline slightly in one direction or the other.
So ultimately the question is: Could there be a better second reference point than the center of the bounding box?
Remark: I am not sure, whether this is the right community for this question or if this question even has a valid answer. Also, I am not sure, which tags to use. Therefore, any advice on more suitable communities and/or tags are very welcome.
I don't think I understand. If the spline is represented in a standard representation such as B-spline, then the parameterization defines the invariant order of the points on the spline. If it is not parameterized, then how is it represented?
@IddoHanniel Thanks for taking your time. I'll try to clarify: A parameterized version of a spline has a certain order. However, there is no reason not to flip the order of the points, i.e. reverse the direction of the spline. The spline would still be principally the same. However, it is important for me, to have a consistent order of points. A simple rule might be: the end-point with the smallest y-coordinate is always the first point of the spline. However, if the spline is rotated, there will be an angle, where the order is flipped. I'd like to find a rule which avoids that.
First, if there is a rotational symmetry in the curve, for example a line segment or an S-shape, then you cannot achieve
your goal since the 180-degree rotated curve is exactly identical to the original.
So, assuming the curve does not have rotational symmetry, what you are asking is an intrinsic property (see e.g., here)
of the curve that will be invariant under rotations.
The first intrinsic property that comes to mind is the curvature profile of the curve, which is invariant to translation and
rotation, and completely describes the shape of the curve.
A practical implementation can be to take the endpoint with the smaller signed curvature as the start/end point of the ordering.
In the degenerate case where both endpoints have the same curvature, you can sample the curvature at a $\delta$-arc-length from both sides and
take the one with smaller curvature of these. Since the curve hasn't got a rotational symmetry and the curvature profile completely describes the curve shape,
at some point there will be a difference in the curvature.
Arc length is also an example of an intrinsic property, which is why the mid-length point you suggested can also be a good starting point, for example as an origin of a coordinate system that has the tangent as its x-axis (and the normal as its y-axis).
However, the direction of this x-axis is not defined, which is why you needed another point.
Choosing the middle point of the bounding box of the curve, as you suggested, can work if the bounding box is aligned with the above intrinsic coordinate system.
This will work in most cases except for the degenerate cases you mentioned where the mid-length point is on (or very close) to the origin.
It should be noted that the center of an axis-aligned bounding box, which is not oriented according to the above intrinsic coordinate system, is not an intrinsic property, since it moves relative to the curve shape as the curve rotates.
There can be curves for which the bounding box mid-point actually crosses the curve as it rotates, which is exactly what you want to avoid.
The example figures below shows such a curve (in the figure it is a polygonal line, but a similar curved spline can be constructed), with its axis-aligned bounding box and its mid-point, under two rotational configurations.
One can see that as the curve rotates the mid-point moves from one side of the curve to the other.
Thanks a lot! I think I can work something out with these suggestions.
|
2025-03-21T14:48:29.934411
| 2020-02-26T13:25:40 |
353602
|
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|
Stack Exchange
|
Proving that a quotient of hypergeometric functions is smaller than a certain function
Im trying to prove that $\forall w \in (0,1), \forall k \in \left(0,\frac{1}{5}\right)$:
$$h_k(w) = \left[\frac{_2F_1\left(\frac{3}{2},1+\frac{1}{k};\frac{1}{2}+\frac{1}{k};\frac{1-w}{1+w} \right) }{_2F_1\left(\frac{1}{2},1+\frac{1}{k};\frac{1}{2}+\frac{1}{k};\frac{1-w}{1+w} \right) } \cdot \frac{1}{1+w}\left(\frac{k}{k-1} + w(1-w) \right)\right] -1 <0 $$
Numerically I can see that the function between square brackets actually fulfills that $\forall w \in (0,1), \forall k \in \left(0,\frac{1}{5}\right)$:
$$ h_k(w)+1 < \frac{1-w}{2} < \frac{1}{2}$$
Of course if you can prove the second identity you obviously have the first (and room to spare). I have been able to prove the following:
$$\lim_{k\rightarrow 0} (h_k(w)+1) = \frac{1-w}{2}$$
But haven't been able to prove that $h_k(w)$ is decreasing in $k$ or anything that might help me. In case you might need it, the way I proved the limit was using the sum definition of the hypergeometric function, taylor expandind around $k=0$ the quotient of Gamma functions to order $k\sqrt{k}$ , and then doing the sums (which are exact), taylor expanding the denominator around $k=0$ again, and multiplying, which grants me (in this analysis I substituted $\frac{1-w}{1+w}$ by $z$):
$$\frac{_2F_1\left(\frac{3}{2},1+\frac{1}{k};\frac{1}{2}+\frac{1}{k};z\right) }{_2F_1\left(\frac{1}{2},1+\frac{1}{k};\frac{1}{2}+\frac{1}{k};z\right) } \sim \frac{1}{1-z} + \frac{kz}{2(1-z)^2} + \mathcal{O}\left(k^2\right) \rightarrow \frac{1}{1-z} \mapsto \frac{1+w}{2w}$$
Which implies
$$h_0(w) +1 = \frac{1-w}{2}$$
|
2025-03-21T14:48:29.934523
| 2020-02-26T14:06:26 |
353603
|
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"Carlo Beenakker",
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|
Stack Exchange
|
Limit of normalized sum of Dirac measures at first $\lfloor p/2\rfloor$ eigenvalues of the sample covariance matrix, with Marcenko-Pastur assumptions?
Let $\lfloor{*}\rfloor$ denotes the nearest integer $\le *$. I'm asking myself the question what's the limit of the part of the empirical spectral distribution corresponding to the first $\lfloor{p/2}\rfloor$ eigenvalues of the sample covariance matrix?
To be more precise, assume that $X=[x_1,...x_n]=[x_{ij}]_{1\le i,j \le n}$ is a $p\times n$ random matrix, where $x_i$'s are individual iid random sample, so that the entries of $X$, namely $x_{ij}$, are iid random variable with mean $0$, variance $1$ and bounded fourth moments.
Next consider: the random measure $Head_{n,p}:=\frac{1}{\lfloor{p/2}\rfloor}\sum_{i=1}^{\lfloor{p/2}\rfloor}\delta_{\lambda_i}$, where $\lambda_1 \ge \lambda_2 \ge \lambda_{\lfloor{p/2}\rfloor}\ge \ldots \ge \lambda_p$ are the eigenvalues of $\frac{1}{p}XX^{T}$.
Then the question is: what's the limit
$$\lim_{n,p \to \infty, p/n \to c \in (0, \infty)} Head_{n,p}?$$
Intuitively, I'd first guess that it should be the Marcenko Pastur distribution with the $c$ replaced by $c/2$, as $\frac{\lfloor{p/2}\rfloor}{n}\to \frac{c}{2}$ as $\frac{p}{n}\to c$, and that one can construct a random matrix $Y \in \mathbb{R}^{ \lfloor{p/2}\rfloor\times n}$ so that all the $\lfloor{p/2}\rfloor$ eigenvalues of $YY^{T}$ correspond to the first $\lfloor{p/2}\rfloor$ eigenvalues of $X$. But then the question would be what can we say about the limit of the following random measure that's a sum of Dirac measures the last $\lfloor{p/2}\rfloor$ eigenvalues? Mathematically that is: what's the limit of
$$Tail_{n,p}:=\frac{1}{\lfloor{p/2}\rfloor}\sum_{i=\lfloor{p/2}\rfloor}^{p}\delta_{\lambda_i}?$$
If I were correct, this should also be the Marcenko Pastur distribution with the $c$ replaced by $c/2$, but I feel it isn't.
Somewhat more generally, one can assume that $p\to \infty$ being a multiple of $q$, i.e. being of the form $p=kq, k\to \infty$ Let $ r, s \in \mathbb{N}$ fixed, and then we can ask for the limit of the random measure:
$$lim_{p,n \to \infty , \frac{p}{n} \to c \in (0,\infty)}\frac{1}{sq-rq}\sum_{i=rq+1}^{sq}\delta_{\lambda_i}$$
Let me denote the Marcenko-Pastur distribution by $\rho(\lambda)$, so that $\rho(\lambda)d\lambda$ gives the probability for an eigenvalue to be in the interval $(\lambda,\lambda+d\lambda)$, in the limit $p\rightarrow\infty$ at fixed $p/n\leq 1$. I order the eigenvalues from large to small, $\lambda_1\geq \lambda_2\geq \lambda_3\cdots \geq \lambda_p\geq 0$ and define $\Lambda_k$ for $1\leq k\leq p$ by
$$\int_{\Lambda_k}^\infty \rho(\lambda)d\lambda=k.$$
The marginal distribution of $\lambda_k$ is sharply peaked at $\Lambda_k$ with a width that vanishes $\propto 1/\sqrt p$. This is known as "spectral rigidity". As a consequence, the desired "head" distribution is simply a truncated Marcenko-Pastur,
$$\rho_{\rm head}(\lambda)=\begin{cases}
2\rho(\lambda)&\text{if}\;\;\lambda<\Lambda_{p/2},\\
0&\text{if}\;\;\lambda>\Lambda_{p/2}.
\end{cases}
$$
The way in which the step wise truncation is smoothed goes beyond the Marcenko-Pastur limit.
Another way to think about this, is to note that the eigenvalue distribution is self-averaging in the limit $p\rightarrow\infty$: you get the MP distribution from a single Wishart matrix, no need to average over many matrices; then, obviously, if you select a subset of the ordered eigenvalues of that single matrix you get the corresponding truncation of the MP distribution. This applies to any subset that contains a number of eigenvalues which scales with $p$, so it also applies to the more general subset in the OP --- just cut out the corresponding interval in the MP distribution.
Here is a little test for $p=10^3$, $n=10^4$, when $\Lambda_{p/2}/n=0.966565$: the plot compares a histogram of the 500 largest eigenvalues (rescaled by $n$) of a randomly generated Wishart matrix with the truncated Marcenko-Pastur. The histogram is a bit noisy (because it's drawn from a single matrix only), but it does follow the truncated MP quite closely.
The marginal distribution of $\lambda_k$ is sharply peaked at $\Lambda_k$ with a width that vanishes $\propto 1/\sqrt p$. Why?
$$\rho_{\rm head}(\lambda)=\begin{cases}
\rho(\lambda)&\text{if};;\lambda<\Lambda_{p/2},\
0&\text{if};;\lambda>\Lambda_{p/2}.
\end{cases}
$$
This doesn't make sense as $p \to \infty$.
Sorry, I undestood nothing....also, math is not written the way you've written, which lacks rigor. I'm not saying your ideas are meaningless, but the way it's presented bears no rigor. Also, please cite links or name theorems.
I do like and appreciate the simulation though...
@Let'stalkmath -- for a derivation of the "self-averaging" property of the eigenvalue density of Wishart matrices, see Sec. 3.2 of Eigenvalue density of linear stochastic dynamical systems: A random matrix approach -- it's really a basic consequence of spectral rigidity, which applies generically to random-matrix ensembles (not only Wishart, also GOE, etc.). The answer to your query as a truncation of the MP distribution then follows directly from this property that a single matrix already produces the MP eigenvalue density.
This is fine except for a factor of 2 (since it needs to integrate to 1....). That is, $\rho_{head}=2\rho$ in the window you mentioned.
Certainly, factor of 2 fixed, thanks.
|
2025-03-21T14:48:29.934982
| 2020-02-26T14:28:15 |
353604
|
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|
Stack Exchange
|
Inequality between Dirichlet and Neumann eigenvalue for Sturm Liouville problem
Consider the following Sturm Liouville problem on an interval $[a,b]$
$$\frac{\mathrm{d}}{\mathrm{d} x}\left[p(x) \frac{\mathrm{d} y}{\mathrm{d} x}\right]+q(x) y=-\lambda w(x) y$$
for given coefficient function $p(x),q(x),w(x)>0$. Here we can assume they are smooth. One can consider two sets eigenvalues
$$0<\lambda_{1}^D<\lambda_{2}^D<\lambda_{3}^D<\cdots<\lambda_{n}^D<\cdots \rightarrow \infty$$
$$\lambda_{1}^N<\lambda_{2}^N<\lambda_{3}^N<\cdots<\lambda_{n}^N<\cdots \rightarrow \infty$$
where $\lambda_i^D$ are Dirichlet boundary condition and $\lambda_i^N$ are Neumann boundary condition. It is quite easy to show $\lambda_i^N\leq \lambda_i^D$ by the variational characterizations of the eigenvalues. However, it seems that under some condition of $p,q,w$, we will have
$$\lambda_{i+1}^N\leq \lambda_i^D$$
I encounter this when reading some papers, and it says this is well known. Does anyone know a good reference for this type of inequality? Who first found this? Do you have a simple proof?
PS. I googled a little bit and found many works on eigenvalue inequality of this type for Laplacian operator on dimension greater than 1. I am just interested in one dimension and here the operator is more general than Laplacian.
I do not know but I have two suggestions. In the 1d case, the book by A. Zettl is a very detailed treatment of Sturm Liouville problems, maybe you find it there. For the Laplacean, Filonov's proof of the inequality is the easiest and could be adapted to more general Sturm-Liouville operators.
Thank you for your comment
|
2025-03-21T14:48:29.935110
| 2020-02-26T14:49:15 |
353606
|
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|
Stack Exchange
|
Link between eigenvalues of a symmetric matrix and a functional space
Let $f_1,\dots,f_n \in L^2(\mathbb{R},\mathbb{R})$ be $n$ mutually orthogonal functions with $\int f^2_i =1$ such that $|\{x \in \mathbb{R} | f_i(x) = 0\}| = 0$ for any $i \in \{1, \dots,n\}$. Does there exist $g \in L^{\infty}(\mathbb{R},\mathbb{R})$ such that all the eigenvalues of the real symmetric matrix $M = \left( \int_{\mathbb{R}} g f_i f_j\right)_{i,j}$ are different ?
The condition that the nodal sets have zero measure gives a lot of freedom to make $M$ vary. For instance we can easily build a $g$ such that all the non-diagonal elements are different to zero, but this is not enough for all the eigenvalues to be different. Actually the statement must be true generically (in the sense that for any starting $g$ for which it's wrong, there exists a perturbation of $g$ such that this becomes true) and it would be even better if one manages to find a proof of it, but I need only the existence of one such $g$ to begin.
Perhaps you want to assume the $f_i$ are not just different but linearly independent? Otherwise, they could all be different scalar multiples of the same function, in which case your matrix has rank $1$ and the eigenvalue $0$ has multiplicity $n-1$.
Yes indeed, I normalized them so that they are linearly independent
Normalizing does not make them linearly independent.
Ok they are orthogonal actually in my problem so now they are linearly independent :p
|
2025-03-21T14:48:29.935246
| 2020-02-26T15:00:22 |
353608
|
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"Asaf Shachar",
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|
Stack Exchange
|
Is this subset of matrices contractible inside the space of non-conformal matrices?
Set $\mathcal{F}:=\{ A \in \text{SL}_2(\mathbb{R}) \, | \, Ae_1 \in \operatorname{span}(e_1) \, \, \text{ and } \, \, A \, \text{ is not conformal} \,\}$, and
$\mathcal{NC}:=\{ A \in M_2(\mathbb{R}) \, | \det A \ge 0 \, \,\text{ and } \, A \text{ is not conformal} \,\}$.
By a non-conformal matrix, I mean a matrix whose singular values are distinct. (i.e. I allow non-zero singular matrices in $\mathcal{NC}$).
Is each connected component of $\mathcal{F}$ contractible in $\mathcal{NC}$?
$\mathcal{F}$ has two connected components, both homeomorphic to an open half-plane with one point removed.
Indeed, $Ae_1 \in \operatorname{span}(e_1)$ and $A \in \text{SL}_2(\mathbb{R})$ imply that
$$ A=\begin{pmatrix} \lambda & y \\\ 0 & \lambda^{-1} \end{pmatrix} \, \, \, \text{for some }\, \lambda \neq 0.$$ $A$ is conformal if and only if $\lambda=\pm 1$ and $y=0$. So, $A$ is not conformal if f $\lambda \neq 1,-1,0$ or $\lambda=\pm 1$ and $y \neq 0$. Thus, one connected component of $\mathcal{F}$ is homeomorphic to
$$\{ 0<\lambda \neq 1\} \times \mathbb{R} \cup \{1\} \times \mathbb{R}\setminus\{0\}.$$
(The second component corresponds to $\lambda <0$.)
Here is what I know about the topology of $\mathcal{NC}$:
Let $\mathcal D=\{ (\sigma_1,\sigma_2) \, | 0 \le \sigma_1 < \sigma_2\}$. Then the map
\begin{align*}
\mu: SO_2\times \mathcal D\times SO_2\to \mathcal{NC}\\
(U,\Sigma,V)\mapsto U\Sigma V^T
\end{align*}
is a $2$-fold smooth covering map*. (i.e. $\mu(U,\Sigma,V)=\mu(-U,\Sigma,-V)$, and this is the only ambiguity in $U,V$ for a pre-image of a given point in $\mathcal{NC}$.
Since $SO_2 \cong \mathbb{S}^1$, and since after identifying antipodal points in $\mathbb{S}^1 \times \mathbb{S}^1$, we get the $2$-torus $\mathbb{T}^2$ again, it follows that $\mathcal{NC} \cong \mathbb{T}^2 \times D$.
*I am not entirely sure regarding the behaviour at the boundary points where $\sigma_1=0$, but I don't think this creates a serious problem.
What is a conformal matrix? I thought fractional linear transformations are conformal but this is probably not what you mean.
A conformal matrix is a nonzero scalar multiple of an orthogonal matrix.
@MoisheKohan, this turns out not to be the OP's definition (see my answer below).
@macbeth: A matrix is a scalar multiple of an orthogonal matrix if and only if all its singular values are equal.
@macbeth Indeed, Moishe Kohan is right. (the two definitions coincide). In the case of $2 \times 2$ matrices, they are not conformal if and only the two singular values are different.
Edited. In the first version of the answer I was assuming that the space in which the contraction was taking place was not $\cal NC$ but the complement to non-conformal matrices in $SL(2,\mathbb R)$. I'll suggest a fix for this now.
Note, that we have a natural continuous map $u: {\cal NC}\to S^1=\mathbb RP^1$. Namely, to each matrix $A$ from ${\cal NC}$ we can associate the following one-dimensional subspace $u(A)\in \mathbb R^2$. Take the matrix $AA^{*}$ and take the eigenspace corresponding to the maximal eigenvalue of $AA^*$ (there will be two distinct eigenvalues since $A$ is not conformal).
So, if we find a closed path $\gamma$ in $\cal F$, such that its image $u(\gamma)\subset S^1$ is not contractible, we are done. How to find such a path is explained in the previous answer to this question, which the path $\gamma(t)$ constructed in the previous answer below. ( I believe that what I suggest works for several reasons but I don't have time to work out all the details now. By the way, it is also funny that $\pi_1(\cal NC)$ seem to be equal $\mathbb Z^2$, moreover it deformation retracts to $T^2$, I believe.)
Previous answer.
It is not contractible. Let us associate to each matrix $A\in SL_2(\mathbb R)$ the following vector $v(A)$. Take an orthogonal matrix $O\in SO_2(\mathbb R)$ such that $OA(e_1)$ is proportional to $e_1$ with a positive coefficient. Then set $v(A)=OA(e_2)$. We get a map to the upper half plane:
$$V:SL(2,\mathbb R)\to \{y>0\}$$
Note that the image of confromal matrices is the point $(0,1)$, and the image of any component $\cal F$ is the complement to $(0,1)$. And so each component can be identified with this puncutred half-plane. Hence it is enough to construct a path in $\cal F$ whose image under $V$ is not contractible in $\{y>0\}\setminus \{(0,1)\}$. This is easy, just take a non-contractible path $\gamma(t)\subset \{y>0\}\setminus \{(0,1)\}$ (that has a non-zero winding number around $(0,1)$), and consider the unique path of matrices $A_t\subset \cal F$ such that $A_t(e_2)=\gamma(t)$.
Hi, after some further thinking I am not so sure that this argument settles the question. The question was whether or not $F$ is contractible inside $\mathcal{NC}:={ A \in M_2(\mathbb{R}) , | \det A \ge 0 , ,\text{ and } , A \text{ is not conformal} ,}$. But the point $(0,1)$ does lie in the image of your map $V:\mathcal{NC} \to M_2(\mathbb{R})$; take $Ae_1=2e_1, Ae_2=e_2$. (We allow matrices outside $\text{SL}_2(\mathbb{R})$ in $\mathcal{NC}$. I think that you have only proved that $\mathcal{F}$ is not contractible inside $\text{SL}_2(\mathbb{R}) \cap \mathcal{NC}$).
In other words, by restricting the domain of $V$ to $\text{SL}_2(\mathbb{R})$ your argument only applies to deformations of $\mathcal{F}$ in $\text{SL}_2(\mathbb{R}) \cap \mathcal{NC}$, while I am allowing arbitrary deformations in $\mathcal{NC}$.
I see Asaf. I read the question too quickly, I thought that you want to contract this in the complement to conformal matrices in $SL(2,\mathbb R)$. Let me think how to fix the argument. I believe this can be done. At last this can be done very easily if you replace $\cal NC$ by ${\cal NC}_+$ of matrices of positive determinant
Asaf, I added some details - not 100% detailed, but this should work I guess. Let me know if you are happy with this.
Thank you for this modification. I am not sure why the path you suggested in the first version should apply in this case as well. Can you say why do you think that this should be easier inside ${\cal NC}_+$ instead of ${\cal NC}$?
Yes, they are contractible in $\mathcal{NC}$, in fact even with $SL$ rather than $GL$. First simultaneously rotate $\mathcal{F}$, so the typical element is sent to
$$
\begin{pmatrix} \cos a & -\sin a \\\ \sin a & \cos a \end{pmatrix}\begin{pmatrix} \lambda & y \\\ 0 & \lambda^{-1} \end{pmatrix}=\begin{pmatrix} \lambda \cos a & y\cos a-\lambda^{-1}\sin a \\\ \lambda \sin a & y\sin a+\lambda^{-1}\cos a \end{pmatrix}
$$
for some small $a$. These are not conformal, since the $(2,1)$-entry is nonzero.
Then simultaneously shrink all these matrices down to the $\lambda=\pm 1$, $y=0$ matrix (depending on the connected component),
$$
\pm\begin{pmatrix} \cos a & -\sin a \\\ \sin a & \cos a \end{pmatrix}.
$$
Thanks. However, by a non-conformal matrix, I refer to a matrix whose singular values are distinct. So $\lambda=\pm 1, y=0$ is exactly the kind of matrices I want to avoid.
|
2025-03-21T14:48:29.935968
| 2020-02-26T15:04:34 |
353609
|
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|
Stack Exchange
|
When is there a subring of the complex numbers surjecting onto a given field of prime characteristic?
To make use of the Lie algebra action of $\mathsf{gl}_2(\mathbb{C})$ to establish a isomorphism in modular representation theory, I would like an answer to this question:
Let $K$ be a field of prime characteristic. When is there a subring $R$ of the complex numbers and a maximal ideal $M$ of $R$ such that $R/M \cong K$?
Clearly no such ring $R$ exists if $K$ has strictly more than $|\mathbb{C}|$ independent transcendental elements. Is this the only obstruction? Is there a reasonably explicit way to construct a suitable $R$ when $K$ is the algebraic closure of $\mathbb{F}_p$?
As a follow-up (which at first I thought I needed, but I now see I can get around by working with $\mathrm{GL}_2(\mathbb{C})$ rather than $\mathrm{SL}_2(\mathbb{C})$), note that if $R/M \cong K$ then the induced map $\mathrm{GL}_d(R) \rightarrow \mathrm{GL}_d(K)$, defined on a $d \times d$ matrix with entries in $R$ by applying $R \twoheadrightarrow R/M \cong K$ to each entry, is a surjective group homomorphism.
Is is true in general that the restriction of the group homomorphism $\mathrm{GL}_d(R) \rightarrow \mathrm{GL}_d(K)$ to $\mathrm{SL}_d(R)$ is surjective onto $\mathrm{SL}_d(K)$, or are there further obstructions?
If $X$ is a transcendence base for $\mathbb C$ over $\mathbb Q$, then the subring it generates is a free commutative ring over $X$. That subring has a surjection onto any ring of cardinality at most $|X|=|\mathbb C|$.
Follow-up question: yes (exercise). Hint: elementary matrices.
Clearly this is very easy for some people. But it genuinely comes up in my research and I think the down-voters are being rather harsh.
@MarkWildon Not shure that downvotes come from easy-solvers (the question of constructibility remains open though) (+1)
@YCor: is the point of your hint that the elementary matrices with entries in $R$ generate $\mathrm{SL}_d(R)$? But this is false when $d=2$ and $R = \mathbb{Q}[X,Y]$, as I think was first shown by Cohn, and is mentioned in the introduction to Suslin's paper, On the structure of the special linear group over polynomial rings.
@MarkWildon no, it's that it generates $SL_d(K)$ for $K$ a field, and the arrow is in the right direction. The map $SL_d(R)\to SL_d(K)$ is already surjective in restriction to the elementary subgroup $E_d(R)$.
@YCor: Thank you, I now see it is not very hard.
An algebraic closure of $\mathbb{F}_p$ is given by $\bar{\mathbb{F}}_p = R/M$ where $R$ is the ring of algebraic integers and $M$ any maximal ideal containing $p$.
The complex numbers have transcendence degree the continuum over $\mathbb Q$ so contain a copy of the field of rational functions in continuum many variables over $\mathbb Q$. This in turn contains the ring of polynomials in continuum many variables over $\mathbb Z$, which surjects onto any ring of cardinality at most the continuum.
Thank you. This answers my first question. Maybe it's too much to ask for a more explicit construction, but I would be interested to know when the subring can be taken inside the ring of algebraic integers of $\mathbb{C}$.
@MarkWildon I think the construction of a transcendence basis of $\mathbb{C}$ over $\mathbb{Q}$ (or one of $\mathbb{R}$ over $\mathbb{Q}$) is not easy (up to my untutored knowledge, only 3 explicit algebraically independent elements are known !).
The subring can be taken inside the algebraic integers iff $K$ is an algebraic extension of $\mathbb{F}_p$. For, let $R$ be the ring of algebraic integers in $\mathbb{C}$. Since any subring $S$ of $R$ is an algebraic extension of $\mathbb{Z}$, the quotient of $S$ by a maximal ideal $m$ is an algebraic extension of $\mathbb{F}_p$ where $(p)=\mathbb{Z}\cap m$. ...
... Conversely, let $K$ be an algebraic extension of $\mathbb{F}_p$. Let $M$ be a maimal ideal of $R$ containing $p$. Then $R/M$ is an algebraic closure of $\mathbb{F}_p$. Hence we can assume $K \le R/M$. Let $\pi: R \to R/M$ and let $S = \pi^{-1}(K)$. Then $L=S/(M \cap S)$. qed
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2025-03-21T14:48:29.936272
| 2020-02-26T15:15:01 |
353610
|
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|
Stack Exchange
|
Asymptotic expansion for sum involving divisor function
Could someone tell me if there is a more precise asymptotic expansion for this sum?
$$\sum_{n\leq x}\frac{d(n)}{n}=\frac{1}{2} \ln^{2}x+2\gamma \ln x+c+O(x^{-\frac{1}{2}}\ln x) $$
$$d(n)=\sum_{a|n}1$$
this is answered at https://math.stackexchange.com/q/471326/87355
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2025-03-21T14:48:29.936333
| 2020-02-26T15:19:32 |
353613
|
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"url": "https://mathoverflow.net/questions/353613"
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Stack Exchange
|
Combinatorially defined effectively closed set
Is there a combinatorially defined, nonempty effectively closed set $Q\subseteq 2^\omega$ such that all members of $Q$ are incomputable?
Combinatorially defined means that the definition of $Q$ does not involve any logic or computability theoretic notion. This excludes those effectively closed sets defined by PA degree, 1-randomness etc.
Effectively closed set means the set $\{\rho\in 2^{<\omega}:[\rho]\cap Q=\emptyset\}$ is comuptably enumerable.
What does "combinatorially defined" mean?
It's explained @NoahSchweber. I'd also be happy to see an algebrically defined one~
It doesn't seem to have anything to do with combinatorics - at present, all you're asking for is a non-"logic-y" example?
You can think of it that way. A "naturally" defined (for non logicians) $Q$.
I think it's a very interesting question. So basically is there something like Hilbert's 10th problem but as a set of reals instead of a set of integers.
The answer is yes.
Let me first describe a general method. Fix any c.e. computably inseparable pair $A$ and $B$. These are computably enumerable sets having no computable separation. There are diverse examples of such pairs of sets. Let $Q$ be the set of sequences $s$ separating $A$ and $B$, that is, for which $s(n)=1$ whenever $n\in A$ and $s(n)=0$ whenever $n\in B$. By assumption, there are no computable elements of $Q$. The set is effectively closed, because we can recognize nonmembers of $Q$ in finite time, simply by waiting for the elements of $A$ and $B$ to emerge and observing whether the membership requirements for $Q$ were followed or not.
Next, let me complete the answer by mentioning that there are numerous examples of computably inseparable pairs, including examples that I would think of as combinatorial.
Let $A$ be the set of Turing machine programs $p$ that halt on empty input with output $0$, and let $B$ be the set of programs that do so with output $1$.
Let $A$ be the set of theorems of PA and $B$ the set of negations of theorems of PA.
Let $A$ be the set of finite Game-of-Life positions that lead eventually to square $1$ turning on before square $2$ (using two fixed squares that makes this example work) and $B$ the set of positions in which $2$ turns on before $1$. This is computably inseparable, because Game-of-Life can simulate Turing machines, and we can pick the squares so that square $1$ corresponds to accept and square $2$ corresponds to reject.
For a particular finite set $\Gamma$ of rational polygonal tiles (or Wang tiles, if you prefer), let $A$ be the set of finite additional possible tiles of a particular form, which admit no tiling of the plane when added to $\Gamma$, and let $B$ be the set of such augmented tile sets of another particular form that admit no tiling when added to $\Gamma$. Since the tiling problem is Turing complete, I can make an example of this form, where the additional tiles of $A$ correspond to halting with output $1$ and those of $B$ correspond to halting with output $0$, but there is no need to talk about Turing computation to describe the actual tiles.
Let $A$ be the set of Post correspondence problems (of a certain form) that admit a solution ending with the symbol $0$ and $B$ the set of such problems admitting a solution ending with the symbol $1$. Since the Post correspondence problem can simulate Turing computation, this example can be created from the first example above.
I think one could create many more examples of a combinatorial nature, simply using the fact that combinatorial questions are often Turing complete and can encode Turing computations.
I suspect that this won't match the OP's criterion of "combinatorially defined" (although it's unclear what that means right now).
It seems perfectly combinatorial to my way of thinking, since it is about the nature of a finitary discrete process. I suppose one could ask what are the most natural computably inseparable sets? Another good example is where A is the set of theorems of PA (or consistent c.e. theory of arithmetic) and B is the set of negations of theorems.
I mean, I would agree, but I find all computability to be combinatorial. :)
So why object then? We can say what counts as combinatorial, and I affirm that the operation of Turing machines counts as combinatorial.
I don't object - I didn't downvote after all - I only meant to get ahead of what I suspect will be the OP's response.
Thanks for responding @JoelDavidHamkins~ But this definition isn't purely combinatorial. I will add more explanation in my question.
The method of the answer is completely general, and works with any computably inseparable pair. So all you need are "combinatorial" computably inseparable sets. But I don't believe that you have a robust concept of what counts as combinatorial, if you exclude all logic and computability theory.
I'm guessing that "naturally occurring in mathematics" can substitute for "combinatorial". In that case, arguably separating A and B is not natural, since what we really want to do is compute A and B themselves.
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2025-03-21T14:48:29.936665
| 2020-02-26T15:23:40 |
353615
|
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|
Stack Exchange
|
Correct reference for a proposition in a paper of Kapranov-Vasserot
In the paper "Kleinian singularities, derived categories and Hall algebras" Math. Ann. 316 (2000) of Kapranov-Vasserot, the authors write in page 569 that the complex $\mathcal{L}'$ (defined in p.568) is quasi-isomorphic to the structure sheaf $\mathcal{O}_{\Delta}$. The reference given for this is the paper of C.M. Ringel, "Hall algebras and quantum groups", Invent. Math. 101 (1990) and in particular Lemma 4.10. However, I am unable to find such a lemma or even why this is true.
Is anyone aware the correct ref. or even better explain why this is true?
Did you try to ask the authors?
There appears to be a production mistake in the published version of the paper. If you look in the copy of the paper on the arXiv, you will find that reference [11] is originally Nakajima's book "Lectures on Hilbert schemes of points on surfaces", whose Section 4 is indeed the correct reference for this material.
Ah, I see. Many thanks for the observation.
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2025-03-21T14:48:29.936779
| 2020-02-26T17:03:28 |
353617
|
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"url": "https://mathoverflow.net/questions/353617"
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|
Stack Exchange
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Reference request : Quotient manifold theorem for Lie groupoid action on a manifold
Let $G$ be a Lie group and $M$ be a smooth manifold. Let $G\times M\rightarrow M$ be a smooth map giving a free, proper action of $G$ On $M$. Then, by quotient manifold theorem, we see that there exists a smooth structure on the quotient set $M/G$ that is well behaved.
Let $\mathcal{G}=(\mathcal{G}_1\rightrightarrows \mathcal{G}_0)$ be a Lie groupoid and $P$ be a smooth manifold.
Let $(\pi:P\rightarrow \mathcal{G}_0, \mu:\mathcal{G}_1\times_{s,\mathcal{G}_0,\pi}P\rightarrow P)$ be a left action of the Lie groupoid $\mathcal{G}$ on the manifold $P$. Further assume that the action is free and proper. Then, we expect the quotient $P/\mathcal{G}_1$ to have a smooth structure as in the case of the quotient manifold theorem. It is true, I have written down some details.
Is there any reference for validity of this result; that is, quotient manifold theorem of Lie group action on manifold extends to Lie groupoid action on manifold?
I have asked this first at MSE https://math.stackexchange.com/questions/3557293/quotient-manifold-theorem-for-lie-groupoid-action-on-a-manifold It has been 4 days, there are 10 views but no response. So, asking here..
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2025-03-21T14:48:29.936884
| 2020-02-26T17:18:49 |
353618
|
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|
Stack Exchange
|
Smoothability of open 4-manifolds
F. Quinn proved that any open topological 4-manifold admits a smooth structure in Ends of maps III: dimensions 4 and 5.
He first proves the generalized annulus conjecture:
Suppose $h:D^j\times \mathbb{R}^{4-j}\to W$ is a homeomorphism of smooth manifolds, smooth on the boundary. If $j=0$ or $1$, then $h$ is isotopic rel boundary and a neighborhood of the end to a map which is smooth on a neighborhood of $D^j\times \{0\}$. When $j=2$, there is an isotopy to a map smooth on either a neighborhood of $D^2\times \{0\}$ after some finger moves to introduce self-intersections, or the image of a neighborhood of $D^2\times \{0\}$ under a topological isotopy (rel $\partial$ and $\infty$).
He then states that "immersion theory interprets the handle result as a calculation of the homotopy groups of classifying spaces" and concludes (Corollary 2.2.3)
$\mathrm{TOP}(4)/O(4)\to \mathrm{TOP}/O$ is 3-connected. Consequently, any 4-manifold has a smooths structure in the complement of a point.
Questions:
Why does the generalized annulus conjecture imply 3-connectedness of $\mathrm{TOP}(4)/O(4)\to \mathrm{TOP}/O$?
Why does 3-connectedness of $\mathrm{TOP}(4)/O(4)\to \mathrm{TOP}/O$ imply that open 4-manifolds are smoothable? I see that it implies that the tangent microbundle of an open 4-manifold admits an $O(4)$-structure, but can the $O(4)$-structure be integrated to a smooth structure on the base (see this post)? What is the argument Quinn is using?
The answer to Question 2 is a standard h-principle argument, see e.g. Theorem 4.4 of Siebenmann's ICM address (https://www.maths.ed.ac.uk/~v1ranick/haupt/siebicm.pdf). An answer to both questions can be found in Section 8.7 of Freedman-Quinn. An easier proof of the corollary is given in Section 8.2 of that book.
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2025-03-21T14:48:29.937030
| 2020-02-26T17:56:28 |
353619
|
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|
Stack Exchange
|
Construction of (general class of) function(s), which sieves out primes, w.r.t. given conditions:
Consider the function $F(x)$ defined in following manner:
$F(n)$ is finite (likely $F(x)\in[0,1]$) if $n$ is prime and zero otherwise:
It has to satisfy following conditions:
(1) $F(x)$ is analytic on $x\in [m,\infty] $ for some positive $m$
(and it is non decreasing at prime values )(optional condition)
(2)$$\lim_{y→∞}|F(x ± iy)|e^{−2πy }= 0$$
uniformly in $x (\geqslant m)$ on every finite interval
(3)$$\int_0^\infty |F(x + iy) − F(x − iy)|e^{−2πy} dy$$
exists for every $x ≥ m$ and tends to zero when $x → ∞$
I've got a function almost satisfying above conditions:
$$F(x)= \sin^2(\frac{π\Gamma(x)}{2x})$$
The above function satisfies condition (1) and (2) but doesn't seem to satisfy the 3rd condition .
Question : How to construct a function that satisfies all above conditions?
Could you write a more specific title? "satisfying the following conditions" is not very useful in a title. Could you start with specifying the domain and range of the function? be more precise that "$F(n)$ is finite"? etc. Currently this looks pretty unclear.
@YCor thank you for the comment , I've edited the question to my best ability . Considering the generality of the question providing domain and range will restrict the problem hugely ; So, I kept it that way .( Also forgive me : my English is not good)
could you remove spacings before punctuation marks? this is wrong typography.
@YCor thank you for the correction , I've edited it.
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2025-03-21T14:48:29.937276
| 2020-02-26T17:58:18 |
353620
|
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Stack Exchange
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Is the inclusion of its 2-skeleton into the walking idempotent homotopy cofinal?
Let $Idem = Idem^{(\infty)}$ be the walking idempotent [1], and let $Idem^{(n)}$ be its n-skeleton. Note that $Idem$ has one nondegenerate simplex in each dimension. Let $\iota_n^m: Idem^{(n)} \to Idem^{(m)}$ be the inclusion. Lurie has shown [2] the following:
If $X$ is a quasicategory, and if $Idem^{(3)} \xrightarrow f X$ is a map, then there exists a map $Idem \xrightarrow g X$ such that $g\iota_1^\infty = f\iota_1^3$.
That is, although a homotopy-coherent idempotent involves infinitely many pieces of coherence data corresponding to the infinitely many nondegenerate simplices of $Idem$, nevertheless all of this data is guaranteed to exist once the first 3 have been found. However, the resulting coherence data might be different from the original data in dimensions 2 and 3 [3].
The proof is a bit involved, and I have not studied it in detail, but what I do understand seems to suggest an affirmative answer to the following
Question: Is the inclusion $Idem^{(n)} \to Idem$ homotopy cofinal for certain $n$ (i.e. is it cofinal in the $\infty$-categorical sense -- depending on how one defines this, it may be necessary to Joyal-fibrantly replace $Idem^{(n)}$ before the question makes sense)? As noted in the comments, this is definitely not true for $n$ odd, nor is it true for $n=0$. Somehow it seems unlikely for $n=2$; hence the title question, which asks this for $n=4$.
I don't believe the inclusion $Idem^{(n)} \to Idem$ is left or right anodyne, so any proof will encounter some complications.
One idea would be to use the usual map $N \to Idem^{(n)}$, where $N \subseteq \mathbb N$ is the graph on which the poset $\mathbb N$ of natural numbers is free -- the 0-cells are natural numbers, and there is a 1-cell from $n$ to $n+1$ for each $n$. For the inclusion $N \to \mathbb N$ is a categorical equivalence, and it's easy to show that $\mathbb N \to Idem$ is homotopy cofinal using Quillen's Theorem A (since the relevant slice categories are just ordinary 1-categories). By composition, $N \to Idem$ is homotopy cofinal, but this factors through $Idem^{(n)}$. By a cancellation property of cofinality, in order to show that $Idem^{(n)} \to Idem$ is homotopy cofinal, it will suffice to show that $N \to Idem^{(n)}$ is homotopy cofinal, which sounds straightforward, since $N$ and $Idem^{(n)}$ are finite-dimensional. But I'm not sure the relevant slice objects are still finite-dimensional...
[1] That is, $Idem$ is the category with one object and one non-identity morphism $i$, satisfying the equation $i^2 = i$. In this question I identify 1-categories with their nerves, which are quasicategories.
[2] HTT <IP_ADDRESS> in the current version. This does not appear in the published version of HTT. It appears in older versions of HA as <IP_ADDRESS>, but was moved to HTT when Lurie rewrote the section on idempotents in HTT.
[3] For example, consider the inclusion $Idem^{(3)} \to \widetilde{Idem^{(3)}}$ given by a Kan fibrant replacement such as $Ex^\infty$. This map is nontrivial on homology, so it cannot actually be extended to a map $Idem \to \widetilde{Idem^{(3)}}$ since this would amount to a trivialization. Thus the extension produced by Lurie's theorem must be changing the image of the 3-cell of $Idem^{(3)}$, at least. In particular, it's not the case that all quasicategories have the right lifting property with respect to $\iota_3^\infty$.
Something is wrong -- a fundamental group / homology computation shows that the weak homotopy type of the $n$-skeleton of $Idem$ is $S^n$ for $n$ odd, and weakly contractible for $n$ even. $Idem$ itself is weakly contractible, and homotopy cofinal functors preserve weak homotopy type, so the inclusion of the $n$-skeleton can only be homotopy cofinal for $n$ even. (This condition is necessary but not sufficient -- e.g. the 0 and 2 -skeleton inclusions are certainly not homotopy cofinal). In particular, the 3-skeleton is not homotopy cofinal in $Idem$, but perhaps the 4-skeleton is.
Another proof of this fact using Quillen cohomology of categories and obstruction theory appears in Example 3.3.11 of https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/topo.12074 (there is a free version on my site, but I see that the arxiv version is not up to date though). Roughly speaking, the point is that $Idem$ has low "Quillen cohomological dimension", and so once you know the first few coherences, the rest of the obstructions are zero.
I don't know what's wrong with the following computation, but the answer is clearly no: if there were a cofinal functor from a finite simplicial set to $Idem$, then any $\infty$-category with finite colimits would have split idempotents, which is not the case (witness finite spaces).
Somewhat surprisingly, this seems to work for even $n>0$, even for $n=2$! That is,
Claim: Let $n \in \mathbb N$. Then the inclusion $Idem^{(n)} \to Idem$ is homotopy cofinal (equivalently, since everything is self-dual: co-cofinal) if and only if $n$ is positive and even.
Proof: We have seen that this can't happen when $n=0$ or $n$ is odd. Otherwise, we verify the hypotheses of the Joyal-Lurie version of Quillen's Theorem A, i.e. we check that the simplicial set $X^{(n)} = Idem^{(n)} \times_{Idem} Dec(Idem)$ is weakly contractible. Here $Dec(Idem)$ is the decalage contruction $Dec(C)_n = C_{n+1}$ where we forget the 0th degeneracy. So in terms of simplices, we have $X^{(n)}_m = Idem^{(n)}_m \times_{Idem_m} Idem_{m+1}$. For $m \geq n+1$, an $m$-simplex in $X^{(n)}_m$ is a string of morphisms in $Idem$ (each either $i$ or $1$) of length $m+1$, such that among the last $m$ morphims in the string, all but at most $n$ are $1$. Any such simplex is a degeneracy of a simplex obtained by deleting one of the copies of $1$. That is, $X^{(n)}$ is $n$-skeletal. Now, the $n$-skeleton of $X^{(n)}$ agrees with that of $X^{(\infty)}$, which is weakly contractible. Therefore, since $n \geq 2$, we have $\pi_1(X^{(n)}) = \tilde H_{\leq n-1}(X^{(n)}) = 0$. So it will suffice to show that $H_n(X^{(n)}) = 0$. There are two nondegenerate simplices of degree $n$: the string $1,i,\dots,i$ (a $1$ followed by $n$ $i$'s) and the string $i,i,\dots,i$ (a string of $n+1$ $i$'s). The boundaries of these (remember that $n$ is even and we are omiting the $\partial_0$ term of the boundary map) are $1,i,\dots,i$ and $i,i,\dots,i$ (where now we have one fewer term in each string) respectively, which are linearly independent. Thus there are no nondegenerate cycles and $H_n(X^{(n)}) = 0$ as desired.
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2025-03-21T14:48:29.937726
| 2020-02-26T18:04:24 |
353621
|
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|
Stack Exchange
|
Conditions to determine sign of real roots
From a delay system, I obtain the following as part of a characteristic equation:
$$f(\lambda) = \lambda - a + be^{-c\lambda},$$
where $a, b,$ and $c$ are positive number and $a<b, ac<1$. My goal is to find the sign of the real part of the root to $f(\lambda) = 0$. Taking $\lambda = x + iy$, I obtain:
\begin{align}
x &= a - be^{-cx}cos(cy)\\
y &= be^{-cx}sin(cy).
\end{align}
I am looking for a condition for $x$ to be negative, but I fail to obtain anything meaningful based on the given constraints (for instance, if $a<0$, then $x<0$). I also narrow down $y \in [-\frac{\pi}{2c},\frac{\pi}{2c}],$ but fail to get something from that as well. Any help would be much appreciated!
All of my simulations show that under the given conditions, $x$ is always negative.
as stated the question is difficult to parse: why do you identify the real part of $\lambda$ with the variable $x$? are you trying to solve $f(x)=0$ ?
@CarloBeenakker Yes. I am trying to solve for $f(x) = 0$.
The argument principle counts the zeros inside a closed contour. Consider applying it to a sequence of contours whose interiors exhaust the region with $x>0$. This way, you could potentially exclude the existence of zeros with $x>0$ (or count them, if they do exist).
Stating a consistent problem might help get meaningful answers. From $\lambda-a+b\exp(c\lambda)=0$, with $\lambda=x+iy$, I get $x=a-b\exp(cx)\cos(cy)$ and $y=-b\exp(cx)\sin(cy)$, but those are not the equations you state. In a problem that is all about signs, it helps to keep them straight.
@MichaelRenardy Thank you for pointing that out. I fixed the equation.
@IgorKhavkine Thank you for the suggestion! I will try to apply the argument principle.
I think your equation is $f(\lambda) = \lambda - a + be^{-c \lambda}$. Let us consider the parametrized family $f_{\varepsilon}(\lambda) = \lambda - a + b e^{-\varepsilon c \lambda}$, where $\varepsilon \in [0,1]$. First of all note that the number of roots $f_{\varepsilon}(\lambda)=0$ in every half-plane $\operatorname{Re}\lambda > \beta$, $\beta \in \mathbb{R}$, is finite and uniformly (in $\varepsilon$) bounded. Suppose we know that for all $\varepsilon$ there is no roots $f_{\varepsilon}(\lambda)=0$ lying on the imaginary axis. Then we can consider a closed contour in $\operatorname{Re}\lambda \geq 0$, which encloses all the roots with $\operatorname{Re}\lambda > 0$ for all $\varepsilon$. From the Rouché theorem it follows that all equations $f_{\varepsilon}(\lambda)=0$ has the same number of roots in this contonur for all $\varepsilon$. It is clear that for $\varepsilon=0$ there is only one root $\lambda = -b + a$, which is negative due to $a < b$. Therefore, $f_{1}(\lambda)=f(\lambda) = \lambda - a + be^{-c \lambda}$ has no roots with $\operatorname{Re}\lambda > 0$.
So now we should provide conditions for the equations $f_{\varepsilon}(\lambda)=0$ to not have purely imaginary roots. I leave this for you.
Thank you very much for your help! Could you please clarify how the family of $f_\epsilon$ satisfies the conditions on Rouche's theorem on the boundary of the closed contour? I apologize if this is trivial.
The dependence of $f_{\varepsilon}$ on $\varepsilon$ is continuous. Thus every $\varepsilon_{0}$ has a small neighbourhood in which $f_{\varepsilon}$ has the same number of roots as $f_{\varepsilon_{0}}$. Indeed, we apply the Rouché theorem to $g:=f_{\varepsilon} - f_{\varepsilon_{0}}$ and $f:=f_{\varepsilon}$; if $\varepsilon$ is sufficiently close to $\varepsilon_{0}$ then on the boundary we have $|g| < |f|$. This proves the statement. Now use the compactness of $[0,1]$ to prove that $f_{\varepsilon}$ has the same number of roots inside the contour for all $\varepsilon$.
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2025-03-21T14:48:29.937996
| 2020-02-26T18:07:09 |
353622
|
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|
Stack Exchange
|
Application of Galois descent
I not understand an assumption done at the beginning of the proof of Rigidity lemma in Moonens and van der Geers book about Abelian variaties (page 12). Here is it:
Question: Why the assumption $k= \overline{k}$ is legit? That is, if we take our varieties $X,Y,Z$ over $k$, build the fiber bundles $X \times_k \overline{k}, Y \times_k \overline{k}, Z \times_k \overline{k}$. Recall, by abuse of notation we mean by $X \times_k \overline{k}$ formally more correctly $X \times_{\operatorname{Spec} \ k} \operatorname{Spec} \ \overline{k}$.
assume we have proved the claim for $X \times_k \overline{k}, Y \times_k \overline{k}, Z \times_k \overline{k}$. How we can descent the claim for initial varieties $X,Y,Z$? I think that the author's had previously a Galoi-descent argument in mind but up to now I failed in working it out correctly.
More precisely, the essence of descent theory is that if we want to verify a property of a morphism, we can also do base change to $\overline{k}$ and verify the same property of the resulted new morphism, see eg here. Unfortunately, here we have a factorization problem and thus we haven't from the starting point a morphism of $k$-varieties which we can pullback to a morphism over $\overline{k}$-varieties. On the other hand, if we simply pullback $f$ and build a factorization over $\overline{k}$ I don't see how this factorization can be descent down through $f$.
There is a candidate morphism. You can pick a k rational point x in X (e.g. identity when X is ab.var) and use this to consider k-morphism $Y\rightarrow X\times Y\rightarrow Z$.
@GTA: The lemma not assumes that $X$ is Abelian variety, thus I'm not sure why we can assume that $X$ has a rational point if $k$ is not algebraically closed (e.g. $x^2+y^2+1$ over $\mathbb{R}$).
If the map factors through the projection, it factors uniquely (the projection is dominant) . So the factorization is automatically galois invariant. In other words factoring is a property, not extra data, so it can be checked after base change.
@user7391733 I was trying to give the simplest reason why it should hold in your context without trying to invoke actual Galois descent. Having a candidate object over $k$ is too simple to be called Galois descent. It is usually rather about whether an object over $\overline{k}$ satisfying compatibility condition actually descends down to an object (not defined a priori) over $k$. This case is the case of descent of morphism of quasicoherent sheaves so it is always effective.
@PhilTosteson: one remark on what you mean by Galois-action on morphisms: When you talking about Galois invariant morphism $h$(in our case the factorization map) between $\bar{k}$-schemes $A,B$, which action you concretly mean? ie how the Galois group "act" here on a morphism $h: A \to B$? Do we have in this context "the action" (ie a canonical one)? The only one with which I'm familar works as follows: Since $h$ incuces a map between $\bar{k}$-valued points $h(\bar{k}): A(\bar{k}) \to B(\bar{k})$ defined by composing $h$ with morphisms $\operatorname{Spec} \ \bar{k} \to A$.
Now a $\sigma \in Gal(\bar{k}/k)$ acts on $f(\bar{k})$ by precomposition. And since $\bar{k}$-valued points are dense this action gives an unique action of $\sigma$ of $h$. Is this the action of Galois group on a morphism you are talking about when you mean a "Galois invariant morphism" or do you mean another one?
|
2025-03-21T14:48:29.938241
| 2020-02-26T18:49:59 |
353626
|
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"Sebastien Palcoux",
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}
|
Stack Exchange
|
Further unexpected behavior involving not √2 and parity
Starting with the fascinating question of Sebastien Palcoux Unexpected behavior involving √2 and parity
let now $\alpha \not= \sqrt 2$ and consider the map
$$f: n \mapsto \left\{
\begin{array}{ll}
\left \lfloor{n/\alpha} \right \rfloor \text{ if } n \text{ even,} \\
\left \lfloor{n \cdot \alpha} \right \rfloor \text{ if } n \text{ odd.}
\end{array}
\right.$$
As in the question of Palcoux let $f^{k+1} := f \circ f^k$ and consider the sequence $(f^k(73) \colon k \in \mathbb{N})$. As noted in the comments to Palcoux questions the sequence with $\alpha = \sqrt 2$ does not seem to be bounded.
I tried some randomly chosen $\alpha = \sqrt {1.9}, \sqrt 3, \sqrt {2.1}, \ldots$. In all these cases the sequence was bounded and ended sometimes with $1$.
Is there any reason why the case $\alpha = \sqrt 2$ is so special?
Thanks! You should have a look to my previous post: https://mathoverflow.net/q/353434/34538 which asks this question in general, and to my next one: https://mathoverflow.net/q/353524/34538 showing that such unboundedness happens (and is proved by Lucia in comment) with the golden ratio (and a family of quadratic algebraic integers), but much more strongly. What is mysterious with √2 is that it provides something intermediate.
|
2025-03-21T14:48:29.938346
| 2020-02-26T19:06:01 |
353627
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353627"
}
|
Stack Exchange
|
Tischler's Theorem on nonvanishing $1$-forms on open manifolds
I have been trying to find a generalized version of the following theorem due to D. Tischler,
Theorem 1. Let $M^n$ be a closed $n$-dimensional manifold. SUppose $M^n$ admits a nonvanishing closed $1$-form. Then $M^n$ is a fiber bundle over $S^1$.
Question. More specifically, is there a version of this theorem for open manifolds either with boundary or noncompact?
Theorem 1 depends strongly on the fact that on a closed manifold $M$ we can find a finite basis for the de Rham cohomology group $H^1 (M, \mathbb R)$ such that given a basis $\gamma_1, \ldots, \gamma_l$ of the free part of $H_1(M, \mathbb Z)$ we have
$$\int_{\gamma_j} \omega_j = \delta_{ij} \quad (Kronecker's \ \ delta)$$
for reference there is [Geometry, dynamics and topology of foliations a first Course, Rojas, Bruno, Scardua - Chapter 8].
So here we should be looking at a manifold with finite cover or $\mathcal l_2$-conditions, but I'm not sure where to find material to work on this. If someone could point me in the right direction, I'd appreciate it. Either a paper or a book on the subject works just fine.
Here is the original paper by Tischler https://core.ac.uk/download/pdf/82577795.pdf.
I am not quite sure what conditions you could impose to make such a generalisation. For example $\mathbb{R}$ has a nowhere vanishing closed $1$-form $dx$.
|
2025-03-21T14:48:29.938466
| 2020-02-26T19:19:26 |
353630
|
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|
Stack Exchange
|
Difference between two largest degrees
Consider a uniform random tournament with $n$ vertices. (Between any two vertices $x,y$, with probability $0.5$ draw an edge from $x$ to $y$; otherwise draw an edge from $y$ to $x$.) Let $S$ be the set of all out-degrees. Let $s_1$ be the largest element of $S$, and $s_2$ the next largest. (If $S$ is a singleton, let $s_2=s_1$.)
Let $c\in (0,1)$ be a constant. What is $\lim_{n\rightarrow\infty}\text{Pr}[s_1-s_2<cn]$?
My guess is that the limit should go to $1$, that is, the two largest out-degrees are close to each other compared to the size of the tournament.
It might be worth parenthesizing the second sentence, or otherwise marking it to indicate that it's just defining what you say in the first sentence; it took me a moment to realize that you weren't starting from a tournament graph and then drawing additional edges...
You guess is correct, assuming that by $x_1$ and $x_2$ you meant $s_1$ and $s_2$.
Indeed, the probability in question is $1-p_n$, where
\begin{equation}
p_n:=P(\exists i\in[n]\ D_i-\max_{j\in[n]\setminus\{i\}}D_j\ge cn),
\end{equation}
where $[n]:=\{1,\dots,n\}$ and $D_i$ is the out-degree of the $i$th vertex. We can write
\begin{equation}
D_i=\sum_{j\in[n]}D_{ij},
\end{equation}
where
\begin{equation}
D_{ij}:=I\{X_{ij}=1\},
\end{equation}
$I\{\cdot\}$ is the indicator function, and $(X_{ij})$ is a random $n\times n$ skew-symmetric matrix whose above-diagonal entries are independent Rademacher random variables (r.v.'s), with $P(X_{ij}=\pm1)=1/2$ if $1\le i<j\le n$.
We need to show that $p_n\to0$ (as $n\to\infty$), for each $c\in(0,1)$. In fact,
\begin{equation}
p_n\le nP(D_1-\max_{j\in[n]\setminus\{1\}}D_j\ge cn)
\le nP(D_1-D_2\ge cn).
\end{equation}
Next,
\begin{equation}
D_1-D_2=D_{12}-D_{21}+\sum_{j=3}^n Y_j,
\end{equation}
where $Y_j:=D_{1j}-D_{2j}$, so that the $Y_j$'s are iid zero-mean r.v.'s, with $|Y_j|\le1$. Also, $D_{12}-D_{21}\le1$. So, by (say) Hoeffding's inequality, for all large enough $n$,
\begin{equation}
p_n\le nP(D_1-D_2\ge cn)\le nP\Big(\sum_{j=3}^n Y_j\ge cn-1\Big)\le ne^{-(cn-1)^2/(2(n-2))}\to0,
\end{equation}
as desired.
The outdegree of every fixed vertex is distributed binomialy with the mean $n/2$ and the variance $n/4$. Hence, the probability that the outdegree deviates from $n/2$ by $cn/3$ at least is extremely small, and by the union bound so is the probability that there is at least one vertex with the outdegree deviating from $n/2$ by $cn/3$. This means that with probability $1-o(1)$, the outdegrees of all vertices are in the range $(n/2-cn/3,n/2+cn/3)$, confirming your conjecture.
@pi66 It means the two largest differ by at most $2cn/3$ with high probability. Actually all the outdegrees are the same within $n^{1/2}\log n$ with high probability, as can be shown by Iosef's method.
|
2025-03-21T14:48:29.938662
| 2020-02-26T19:57:53 |
353633
|
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|
Stack Exchange
|
Rank of Vandermonde matrices
Consider a Vandermonde matrix
$$V = \begin{bmatrix}
1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \\
1 & x_2 & x_2^2 & \cdots & x_2^{n-1} \\
& & \vdots \\
1 & x_n & x_n^2 & \cdots & x_n^{n-1}
\end{bmatrix}$$
It is a simple exercise to prove that if the rank of the Vandermonde is $< n$, then the determinant is $0$ and not all of the $x_i$ can be distinct. But is the following more fine-grained statement true?
Question: If a Vandermonde matrix has rank $k$, then does the set $\{ x_1,\cdots,x_n \}$ contain exactly $k$ distinct elements?
A similar statement which can be used to answer your question is quoted in the Wikipedia entry for Vandermonde matrices; so presumably this result is well-known.
@WillieWong I see! Thanks for the reference. I was not aware of this result.
See also (for a more general, noncommutative setting): T. Y. Lam, A general theory of Vandermonde matrices, Expositiones Mathematicae 4 (1986), pp. 193--215.
If ${x_1,\ldots,x_n}$ has exactly $k$ distinct elements, then $V$ has a $k \times k$ invertible submatrix, and has only $k$ distinct lines, so the rank must be $k$.
@FrançoisBrunault That is indeed true, but my question was about the converse statement: does the rank of $V$ being equal to $k$ imply that there are exactly $k$ distinct elements in ${ x_1,\cdots,x_n }$?
Well what I wrote says that the rank is equal to the cardinality of ${x_1,\ldots,x_n}$, which is what you want.
Oh, yes that is correct. My mistake.
|
2025-03-21T14:48:29.938805
| 2020-02-26T20:15:49 |
353635
|
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|
Stack Exchange
|
Application of higher categories in algebra
Higher categories and derived algebraic geometry are relatively new areas and probably fewer people are working on them compared to the majority of topologists or geometers. I believe higher categories should have great power in producing new results in algebra (K-theory, algebraic topology, algebraic geometry). What are some important applications of higher categories in these areas?
'Must one obtain a very deep knowledge of AG or AT to understand the applications of higher topos properly?' I doubt that, since the applications seem to be of a more formal nature. I think most working algebraic geometers don't know what a topos is (even Serre admitted that he doesn't know), and I think this says much regarding the importance of topos theory in algebraic geometry.
I heard higher topos theory and derived algebraic geometry are crucial in String field theory, a branch of mathematical physics.
In String field theory, I've seen little or know typos theory. Drived' stuff is very useful but need not be in AG. See Higher Structures' If it doesn't endanger you, where are you currently
Yes, the journal
This paper https://arxiv.org/pdf/1102.1150.pdf shows how the derived geometry (or homotopical) approach allows to get functoriality "for free" in the construction of perfect obstruction theories.
The Cesnavicius-Scholze purity theorem https://arxiv.org/abs/1912.10932 uses heavily derived methods like animated commutative rings. If this counts as an application of "Higher categories and derived algebraic geometry", may be a matter of taste.
There are tons of such applications. For example, less than two years ago, they were used to prove the Redshift Conjecture, a statement about iterated algebraic $K$-theory. I previously wrote an answer describing the work that went into that proof (of many authors, over many years), and how $\infty$-categories were used.
Another great application of $\infty$-categories was the even more recent disproof of the Telescope Conjecture in algebraic topology, by Robert Burklund, Jeremy Hahn, Ishan Levy, and Tomer Schlank (some of the same authors as the above work). This conjecture was the last of the Ravenel Conjectures from 1984, regarding the structure of chromatic homotopy theory. The telescope conjecture was proved early on for $n=1$ and now we know it's false for $n\geq 2$, which means that the homotopy groups of spheres are even more complicated than was hoped in the 1980s. Quanta magazine covered this. What's exciting to me is that the method of disproof continues the development of our computational power in algebraic $K$-theory, and suggests we will continue to learn cool things in stable homotopy theory with these new computational tools.
In both cases, I think it's safe to say that the use of $\infty$-categories served as a powerful organizational tool, and using the yoga of $\infty$-categories severely reduced the technical hurdles required in applying the main idea to the problem at hand.
If you're willing to broaden what you consider to be "applications of higher categories," let me mention that model categories have also had major applications in algebra and algebraic topology, and experts in $\infty$-categories often tell me that these advances could be re-proven with $\infty$-categories if necessary. One that springs to mind is the resolution of the Kervaire Invariant One problem by Hill, Hopkins, and Ravenel. That's certainly algebraic topology, but perhaps more geometric than you would like. Another example is Voevodsky's Fields Medal winning work proving the Milnor Conjecture. He used a model category structure on motivic spectra, but this work could almost certainly be redone using the corresponding $\infty$-category structure (though, I don't know for sure that anyone has written down such a treatment). See also this thread and this one (which includes a discussion of Vandiver's Conjecture). One last example is the resolution of the Flat Cover Conjecture in homological algebra, via cotorsion pairs and model categories (again, the model categories could be removed today, but they were important in the moment based on what people understood at the time).
|
2025-03-21T14:48:29.939106
| 2020-02-26T21:12:16 |
353637
|
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|
Stack Exchange
|
Example of a projective bimodule with isomorphic left and right duals
What is an example of a non-free finitely generated $R$-bimodule $M$ satisfying
i) $M$ is projective as both a left and right $R$-module
ii) the right dual $\mathrm{Hom}_R(M,R)$ and the left dual $_R\mathrm{Hom}(M,R)$ are isomorphic as bimodules,
where $R$ is a noncommutative unital algebra defined over a field $k$ with non-zero characteristic.
Why does $M=R$ not work?
I guess I want a non-trivial example.
You mean that $M$ is not free?
yes, I have put this in the question
Just a minor rant about terminology. $M=R$ is not free as a bimodule. Also, a bimodule that is projective on the left and on the right is not necessarily a projective bimodule.
Take $G$ a nonabelian finite group of size coprime to $p$, and $k$ a field of characteristic $p>0$. Then $k[G]$ is semisimple (Maschke's theorem), so every module on left or right is projective.
In particular, take $\underline{k}$ to be the trivial representation of $G$. So this is a bimodule projective on both sides, and $\mathrm{Hom}_G(\underline{k}, k[G])={_G\mathrm{Hom}}(\underline{k}, k[G])=\underline{k}$, as bimodules (both of these being naturally the same subspace of $k[G]$, i.e. $\langle\sum_{g \in G}g\rangle$).
... This satisfies all the criteria of the question; somehow, however, this still does not feel as if it should count for a non-trivial example.
In fact, for any group algebra $k[G]$ of a finite group, semisimple or not, and for any bimodule $M$, the left dual and right dual are isomorphic. And there are plenty of natural bimodules that are projective on both sides: e.g., $k[G]\otimes_{k[H]}k[G]$ for a subgroup $H\leq G$.
|
2025-03-21T14:48:29.939362
| 2020-02-26T21:25:58 |
353638
|
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Stack Exchange
|
Deriving asymptotic variance of generalized estimating equation estimator (GEE)
As well known to us, K.Y. Liang and S. Zeger proposed GEE for longitudinal data analysis in their famous paper[1]. At the appendix of the paper, authors show the proof of Theorem 2. I tried to reproduce that proof following their idea, but when I tried to derive $A_{i}$, I failed. Here, I want to add two comments:
"... $\sum A_{i}/K$ converges as K $\rightarrow \infty$ to $-\sum D_{i}^{T}V_{i}^{-1}D_{i}/K$."
I don't think it involves any asymptotics. That is, whatever $K$ is, it holds still. (Do you agree ?)
$A_{i} = \partial U_{i}\{\beta, \alpha^{*}(\beta)\}/\partial \beta$, as stated above, $A_{i}$ should be equal to $-D_{i}^{T}V_{i}^{-1}D_{i}$, however, I failed to derive it. Note that, $U_{i} = D_{i}^{T}V_{i}^{T}S_{i}$, if we fix $D_{i}$ and $V_{i}$ as constant, that is, not related with $\beta$, then $\partial S_{i}/\partial \beta = - A_{i}\Delta_{i}X_{i} = - D_{i}$. But, $D_{i}$ and $V_{i}$ are all related with $\beta$. So, I feel very confused about that!
GEE has been popularly used in longitudinal data analysis and has been further studied in more complicated circumstances by many statisticians. So, the results in Liang's paper should be correct, otherwise, there should be someone pointing it out. Do you have any idea about the proof ?
Thank you very much!
[1]: K.Y. Liang, S. Zeger, longitudinal data analysis using generalized linear models, https://academic.oup.com/biomet/article/73/1/13/246001.
You are right, a sentence like this, "$\sum A_i/K$ converges as $K\to\infty$ to $-\sum D_i^{T}V_{i}^{-1}D_i/K$", cannot possibly have a meaning. Perhaps, the authors meant here someting like this: "$\sum A_i/K+\sum D_i^{T}V_{i}^{-1}D_i/K\to0$ as $K\to\infty$".
@IosifPinelis Thank you Professor. It's possible.
The key point is Law of Large Numbers.
Answer:
$\partial \frac{1}{K}\sum A_{i}/\partial \beta = \frac{1}{K}\sum \partial A_{i}/\partial \beta = \frac{1}{K}\sum \{[\partial(D_{i}^{T}V_{i}^{-1})/\partial \beta \times S_{i}] + [D_{i}^{T}V_{i}^{-1} \times \partial S_{i}/\partial \beta]\}$.
Let $E_{1} = \frac{1}{K}\sum [\partial(D_{i}^{T}V_{i}^{-1})/\partial \beta \times S_{i}]$, apply LLN to $E_{1}$, we have $E_{1} \rightarrow \mathbb{E}[\partial(D_{i}^{T}V_{i}^{-1})/\partial \beta \times S_{i}] = \partial(D_{i}^{T}V_{i}^{-1})/\partial \beta \times \mathbb{E}[S_{i}] = 0$, as $K \rightarrow \infty$.
Let $E_{2} = \frac{1}{K}\sum [D_{i}^{T}V_{i}^{-1} \times \partial S_{i}/\partial \beta] = -\frac{1}{K}\sum D_{i}^{T}V_{i}^{-1}D_{i}$.
Done.
|
2025-03-21T14:48:29.939527
| 2020-02-26T21:36:50 |
353641
|
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|
Stack Exchange
|
Thickness and hierarchical hyperbolicity
Thick metric spaces were introduced by Behrstock, Drutu and Mosher, see here. Hierarchically hyperbolic spaces were introduced by Behrstock, Hagen and Sisto, see here.
I've heard that it is open whether hierarchically hyperbolic groups are thick. Of course, one has to restrict to hierarchically hyperbolic groups that are not non-elementary relatively hyperbolic (NERH for short), for it was proved in Behrstock, Drutu, Mosher that thick groups cannot be NERH. See also there, where Levcovitz proved that thick groups have trivial Floyd boundary, which cannot happen for NERH groups, since the Floyd boundary covers the Bowditch boundary.
Question : Do we know if this is true among all known examples of hierarchically hyperbolic groups (that are not NERH) ?
Note for example that we know that mapping class groups and Artin groups that are not NERH are indeed thick.
Special emphasis on CAT(0) cube complexes. As far as I know, we still don't know if all cubical groups are hierarchically hyperbolic, although to my knowledge, it's likely to be true. On the other hand, do we know if all (not NERH) cubical groups are thick ?
To my knowledge, the question is even open for cocompact special groups, which are hierarchically hyperbolic. (Notice that the relative hyperbolicity among cocompact special groups is studied in arXiv:1709.01258; my guess is that a positive answer could be proved by improving some of the arguments.)
This is not an answer, just some sketchy thoughts that are too long for the comment box. I and some other HHS enthusiasts are very interested in this question being answered; we've tried a fair bit and have set it aside, so I don't think they'll mind me trying to recall what some of the strategies and issues are.
It's indeed open for cocompact special groups, as far as I know. (Part of the motivation for the question comes from Coxeter groups, which are known to be hyperbolic relative to thick subgroups [Theorem VII here], and many of which are virtually compact special.)
There's a very similar question: for the classes of groups in question (HHG, cocompactly cubulated, virtually compact special), is it true that any group $G$ in the class is hyperbolic relative to a collection of subgroups, each of which has at most polynomial divergence (we're allowing $G$ to be hyperbolic relative to itself)?
In particular, does super-polynomial divergence imply exponential divergence for such $G$? (A nontrivially relatively hyperbolic group has at least exponential divergence [Theorem 6.13 here].)
It's also interesting to ask whether $G$ is hyperbolic relative to NRH subgroups (whether because the peripheral subgroups are thick, or NRH for some other reason).
Here is a naive approach that has been tried a couple of times (once for cubical groups, once for HHG) unsuccessfully. I think it's worth sketching because it's probably the first thing one might try, so it might save someone some work to see what the problems with it are. (Maybe they are surmountable.)
The idea is to build candidate thick/NRH subspaces inductively by "brute force", show that this construction terminates, and then show that the resulting subspaces give a valid peripheral structure.
More precisely: let $G$ act geometrically on a proper CAT(0) cube complex $X$, or a proper hierarchically hyperbolic space $X$. In either case, either $X$ is already hyperbolic, or there is a nontrivial product region $P$. (In the cubical case, $P$ is a convex subcomplex decomposing as the product of two unbounded CAT(0) cube complexes. In the HHS case, $P$ is a hierarchically quasiconvex subspace admitting a coarse-median-preserving quasi-isometry to the product of two unbounded hierarchically hyperbolic spaces. The HHS fact follows from Corollary 2.16 here.)
Let $\mathcal P_0$ be the ($G$--invariant) set of such product regions. Call $P,P'\in\mathcal P_0$ "elementary equivalent" if they have unbounded coarse intersection. Taking the transitive closure gives an equivalence relation on $\mathcal P_0$. For each equivalence class, one can take the union $U$ of the subspaces in the class. These $U$ are your candidate thick-of-order-at-most-$1$ pieces.
Now iterate this process: take the set of such $U$, impose the same equivalence relation, and make candidate thick-of-order-at-most-$2$ pieces, etc.
(If I remember right, the equivalence relation used at the $n^{th}$ stage, $n\geq 1$, is that $P,P'\in\mathcal P_{n-1}$ are elementary equivalent if their hierarchically quasiconvex hulls have unbounded coarse intersection. These are described nicely by Russell-Spriano-Tran in Section 5 of this paper; in the cubical case, you use cubical convex hulls.)
Suppose the process terminates, i.e. at some point you have a $G$--invariant collection $\mathcal P_n$ of subspaces, no two of which have hierarchically quasiconvex hulls with unbounded coarse intersection.
At this point, you should:
(1) Verify that each $P$ is "quasiconvex" in the sense appropriate to the category you're working in. In the cubical case, you should check that $P$ is at finite Hausdorff distance from its cubical convex hull; in the HHS case, you should check that $P$ is \emph{hierarchically quasiconvex} --- Proposition 5.11 in this paper by Russell-Spriano-Tran is probably the best way to go about this.
(2) Hope that the stabiliser of each $P\in\mathcal P_n$ acts coboundedly. ($G$ acts on $X$ coboundedly, $\mathcal P_n$ is $G$--invariant; you want to use properness of $X$ and boundedness of the coarse intersections between the elements of $\mathcal P_n$ to show that only finitely many elements of $\mathcal P_n$ intersect any given ball.)
Already there is something tricky here.
(3) Verify that adding combinatorial horoballs over the elements of $\mathcal P_n$ in $X$ gives you something hyperbolic.
I think one approach was to verify hyperbolicity by finding a hierarchically hyperbolic structure with no interesting "product regions".
(4) Hope that, more or less by construction, each $P\in\mathcal P_n$ is thick of order at most $n$.
At that point, you'd have what you want.
Difficulties include:
(a) It's not remotely clear that this process terminates. (But I think it's worth nailing down whether or not it does).
(b) If it doesn't terminate, one can still hope to be able to pass to some sort of limit, and find a collection $\mathcal P_\infty$ of subspaces, all with bounded coarse intersection, to get candidate peripherals in a relatively hyperbolic structure. If this works, one doesn't get that the peripherals are thick, but the hope is that they are at least not relatively hyperbolic.
(c) If I remember correctly, one big difficulty is that, at each (finite) stage of the induction, one might have to pass to larger and larger neighbourhoods to get something convex/hierarchically quasiconvex, and this creates problems if the process doesn't terminate.
More vaguely, at each step, there are various constants to control, and they blow up if the process doesn't terminate.
There was a quite different approach (for CAT(0) groups) discussed by many people at the AIM in 2016. There was some subsequent work on it and some useful ideas, but I'll have to check with those involved to see if it's okay to mention the idea/subsequent developments here --- it's possible that one or more of them are still actively working on it, although I'm not.
So, the short answer, is that it's a very interesting question for which the naive approach is a bit of a mess.
My feeling is that if there's a counterexample, then there's a counterexample where $G$ is $\pi_1$ of a CAT(0) square complex.
Thank you very much for this amazing answer. I'm waiting a bit before accepting it, but it seems to me like a complete answer, although not positive.
|
2025-03-21T14:48:29.939980
| 2020-02-26T22:34:15 |
353644
|
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|
Stack Exchange
|
Asymptotic of an integral
Let
\begin{equation*}
V(x) = -\big(2-\sin(2\pi x) - \sin(2\pi \sqrt{2}x)\big)^\gamma
\end{equation*}
for some $\gamma \in (0,1]$. Define for each $r<0$ the number
$$a_r = \min\{a>0: V(a) = r\}$$
I want to find the asymptotic behavior of the following integral
$$ s_r = \int_0^{a_r}\frac{dx}{\sqrt{2(r-V(x))}}$$
as $r\rightarrow 0^-$. One thing I have in mind is, we can show that
$$ \frac{C_1}{t^{2\gamma}} \leq \min_{[0,t]}|V(x)| \leq \frac{C_2}{t^{2\gamma}}$$, therefore as $|r| = \min_{[0,a_r]} |V(x)|$ one can deduce
$$ C_1^{\frac{1}{2\gamma}}|r|^{-\frac{1}{2\gamma}} \leq |a_r|\leq C_2^\frac{1}{2\gamma} r^{-\frac{1}{2\gamma}}$$
but I failed to proceed further for $s_r$.
This question arises from my attempt to understand the following Hamiltonian
$$H(x,p) = \frac{|p|^2}{2} +V(x)$$
Fixing a parameter $\varepsilon$, I am interested in the behavior of
$$ \varepsilon \eta(\varepsilon^{-1})$$
as $\varepsilon \rightarrow 0^+$ where $\eta$ solves the Euler-Lagrange equation
$$ \ddot{\eta}(s) = -V'(\eta(s)), \qquad \text{on}\; (0,\varepsilon^{-1})$$
and $\eta(0) = 0$. Using the conservation of energy
$$ \frac{|\dot{\eta}(s)|^2}{2} + V(\eta(s)) = r$$
we can deduce the ODE
$$ |\dot{\eta}(s)| = \sqrt{2(r-V(\eta(s)))}, \qquad{on}\;(0,\varepsilon^{-1})$$
The value $s_r$ above is the time it takes for $\eta_r$ to reach to $a_r$, when $r<0$. It is easy to see that, for a fixed $r>0$ we have
$$ \varepsilon \eta_r(\varepsilon^{-1}) \leq \varepsilon a_r \rightarrow 0$$
as $\varepsilon \rightarrow 0$. The question is, can I do that uniformly, in the sense that either
$$ |\varepsilon \eta_r(\varepsilon^{-1})| \leq C \varepsilon ^\alpha $$
or
$$ |\varepsilon \eta_r(\varepsilon^{-1}) - M(r,\varepsilon)| \leq C \varepsilon ^\alpha $$
for some $M_r \rightarrow 0$ as $\varepsilon\rightarrow 0$?
(i) Where does this problem arise? (ii) Did you use the continued fraction for $\sqrt2$ to get $ \frac{C_1}{t^{2\gamma}} \leq \min_{[0,t]}|V(x)| \leq \frac{C_2}{t^{2\gamma}}$? (iii) I think there are typos in your bounds on $a_r$ (which is a generalized inverse of $V(a)$).
(i) This arises from an attempt to quantify the long time average of the solution to the ODE
$$ \begin{cases}
|\dot{\eta}(s)| &= \sqrt{2(r-V(\eta(s)))}, s>0\
\eta(0) &= 0.
\end{cases}$$
(ii) I used Diophantine Approximation of $\sqrt{2}$ to show that inequality (I think it relates to the continued fraction for $\sqrt{2}$ as you said)
(iii) Thanks for that, I did have a typo.
We can see that (numerically or analytically), for a.e. $r<0$ the solution above $\eta(s)$ is periodic, the point here is I am trying to quantify how large is the size of the orbit changes in terms of the time $\eta(s)$ needs to reach to these maximum points.
Thank you for your answers. It is still unclear to me where your ODE arises.
It arises from the conservation of energy $\frac{|\dot{\eta}|^2}{2} + V(\eta) = r$, where $\eta$ solves the Euler-Lagrange equation $\ddot{\eta} = -V'(\eta)$. I am looking at the the minimizer with negative energy.
I guess my questions really were about the function $V$: where does it arise?
So classical mechanic Hamiltonians are well studied for mostly periodic potential, I am considering quasi-periodic potential, and this is the simplest one. The power $\gamma$ is just for technicality.
|
2025-03-21T14:48:29.940202
| 2020-02-26T22:58:18 |
353648
|
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|
Stack Exchange
|
Asymptotic of an area integral
I have the following integral
$$
I(\varepsilon) = \iint_D \frac{\sqrt{1+|\nabla h(u,v)|^2}}{[(h(u,v)+\varepsilon)^2+u^2+v^2]^2} du dv,
$$
where $h$ is a smooth function with
$h(0,0)=0 = h_u(0,0) = h_v(0,0)$, $D$ is a disk centered at the origin.
It seems like that the asymptotic of $I(\varepsilon)$ depends only on the first order behavior of $h$ at the origin, so the dominant term of $I(\varepsilon)$ is the same as that with $h\equiv 0$ and $$I(\varepsilon) \sim \frac{\pi}{\varepsilon^2}, \quad \varepsilon \rightarrow 0.$$
Any suggestion on how to prove/disprove it?
Does "smooth" mean $C^1$?
I was thinking $C^\infty$, or even just $h(u,v) = \kappa_1 u^2 + \kappa_2 v^2$ as a start.
Then split the domain of integration into $u^2+v^2\le\varepsilon^{4/3}$ and the complement.
Do you mean that the dominant asymptotic of $I(\varepsilon)$ comes from restricting the integral to the $\varepsilon^{4/3}$ disc? Why is it so? And would it help proving the $\pi/\varepsilon^2$ asymptotic? Thanks.
$\varepsilon^{2/3}$- disk, that is. Iosif used $3/4$ instead of my $2/3$, but the logic is the same.
$\newcommand{\ep}{\varepsilon}$
This is indeed a matter of splitting the integral. In polar coordinates, for some real $R>0$,
\begin{equation}
I(\ep) = \int_0^{2\pi}(J_t(\ep)+K_t(\ep))\,dt,
\end{equation}
where
\begin{equation}
J_t(\ep):=\int_0^{r_*}\frac{\sqrt{1+|\nabla h|^2}}{\big((h+\ep)^2+r^2\big)^2}\,r\,dr,
\end{equation}
\begin{equation}
K_t(\ep):=\int_{r_*}^R\frac{\sqrt{1+|\nabla h|^2}}{\big((h+\ep)^2+r^2\big)^2}\,r\,dr,
\end{equation}
$h:=h(r\cos t,r\sin t)$, $\nabla h:=\nabla h(r\cos t,r\sin t)$,
$r_*=r_*(\ep)>0$ varies with $\ep\downarrow0$ so that
$$\ep<<r_*<<\ep^{1/2}$$
(e.g., one may take $r_*=\ep^{3/4}$),
$a\ll b$ means $|a|=O(b)$, and $a<<b$ means $|a|=o(b)$.
We have $h\ll r^2$ and $|\nabla h|\ll r$, so that (as $\ep\downarrow0$) uniformly in $r\in[0,r_*]$ we have $h\ll r_*^2<<\ep$, $h+\ep\sim\ep$, $(h+\ep)^2+r^2\sim\ep^2+r^2$, $|\nabla h|<<1$, and $\sqrt{1+|\nabla h|^2}\sim1$. So,
\begin{equation}
J_t(\ep)\sim\int_0^{r_*}\frac{r\,dr}{\big(\ep^2+r^2\big)^2}
=\frac1{2\ep^2}\,\int_0^{r_*^2/\ep^2}\frac{ds}{(1+s)^2}
\sim\frac1{2\ep^2}\,\int_0^\infty\frac{ds}{(1+s)^2}
=\frac1{2\ep^2}. \tag{1}
\end{equation}
On the other hand,
\begin{equation}
K_t(\ep)\ll\int_{r_*}^\infty\frac{r\,dr}{(r^2)^2}\le\frac1{r_*^2}<<\frac1{\ep^2}.
\end{equation}
Thus, indeed
\begin{equation}
I(\ep)\sim\frac\pi{\ep^2}.
\end{equation}
It seems like your claim $J_t(\varepsilon) \sim \int_0^{r_\ast} \frac{r dr}{(\varepsilon^2 + r^2)^2}$ follows solely from the assumption that $r_\ast(\varepsilon) << \varepsilon^{1/2}$ (and that $h$ has a vanishing linearization at the origin.) I could not follow the argument, but now also think that it cannot be true: if all I have is an upper bound on $r_\ast$ , then I can make it zero and then $J_t$ is also zero. (Put differently, the $1/(2\varepsilon^2)$ asymptotic cannot follow just from an upper bound for $r_\ast$.) Am I missing anything?
p.s. The proof for $K_t$ growing strictly slower than $1/\varepsilon^2$ is very clear, and uses only the lower bound for $r_\ast$.
@TYp : I have added details on the first asymptotic equivalence, $\sim$, in (1). In that equivalence, you cannot set $r_=0$, because then the equivalence would become the nonsensical "equivalence" $0\sim0$. More importantly, the lower bound on $r_$ given by $\varepsilon<<r_*$ is needed for the second asymptotic equivalence in (1).
Now I see the lower bound is used and the doubt is gone. Thanks. The technical issue I have is how that "interchange of integral and asymptotic" in the first equivalence can be justified. For simplicity, if $h(u,v)$ is the radially symmetric $1/2(u^2+v^2)$. Then $J_t(\varepsilon)$ is exactly $\int_0^{\varepsilon^{3/4}} \frac{\sqrt{1+r^2} r dr}{[(r^2+\varepsilon)^2+r^2]^2}$. And the claim is that for small $\varepsilon$ this is equivalent to $\int_0^{\varepsilon^{3/4}} \frac{r dr}{[\varepsilon^2+r^2]^2}$. I am trying to figure out the (elementary?) missing lemma.
@TYp : When people write (as I did) "for $r\le r_$ as $\epsilon\to0$", they of course mean "uniformly in $r\in[0,r_]$ as $\epsilon\to0$" (which is how I have now rewritten that). So, the first equivalence in (1) follows by the general fact that, if $f_\epsilon(r)\sim g_\epsilon(r)>0$ uniformly in $r\in[0,r_]$ (as $\epsilon\to0$), then $\int_0^{r_}f_\epsilon(r),dr\sim \int_0^{r_*}g_\epsilon(r),dr$. This fact should be an easy exercise for you.
I have no doubt the proof works, but the only way I can make sense out of "$f_\varepsilon(r) \sim g_\varepsilon(r)$ uniformly in $r\in [0,r_\ast]$" is to show $\max_{r \in [0,r_\ast]} | f_\varepsilon(r) - g_\varepsilon(r) | \leq C \varepsilon^{-2}$. This would definitely imply the conclusion, but the uniform bound is tedious to show. Maybe I am missing a easier trick.
@TYp : No, you don't need to also show that $\max_{r\in[0,r_]}|f_\epsilon(r)-g_\epsilon(r)|\le C\epsilon^{-2}$.
You have $f_\epsilon(r)\sim g_\epsilon(r)>0$ uniformly in $r\in[0,r_]$ (as $\epsilon\to0$).
Previous comment continued: This means that $\forall\delta>0$ $\exists\epsilon_\delta>0$ $\forall\epsilon\in(0,\epsilon_\delta)$ $\forall r\in[0,r_]$ $1-\delta\le f_\epsilon(r)/g_\epsilon(r)\le1+\delta$ or, equivalently, $(1-\delta)g_\epsilon(r)\le f_\epsilon(r)\le(1+\delta)g_\epsilon(r)$, whence $(1-\delta)\int_{[0,r_]}g_\epsilon(r),dr\le \int_{[0,r_]}f_\epsilon(r),dr\le(1+\delta)\int_{[0,r_]}g_\epsilon(r),dr$. Thus, $\int_{[0,r_]}f_\epsilon(r),dr\sim\int_{[0,r_]}g_\epsilon(r),dr$. This is it.
Indeed this is much better. Thanks for the clarification.
|
2025-03-21T14:48:29.940531
| 2020-02-26T23:18:44 |
353650
|
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|
Stack Exchange
|
Example of a nonconvex Chebyshev set in a metric space with continuous projection?
Question: Is there an example of a nonconvex Chebyshev set $S$ in a metric space $(X,d)$ whose projection map is continuous?
For convexity to be well-defined, we need to assume that $X$ is a vector space, but not necessarily a normed space. The projection map is defined as
$$
P_Sx
= \text{argmin}_{y\in S} d(x,y).
$$
A Chebyshev set is a set $S$ such that $P_Sx$ is unique and hence well-defined for all $x\in S$.
It is a classical result that if $X$ is a Banach space and $d$ is the Banach space norm, then continuity of $P_S$ is equivalent to convexity of $S$. Thus no example exists in a Banach space.
1 Wulbert, D. E. (1968). Continuity of metric projections. Transactions of the American Mathematical Society, 134(2), 335-341.
Erm... What is the relation between the metric and the linear structure in the question?
@fedja There is none. The linear structure is needed only to ensure the notion of “nonconvexity” is well-defined.
Then nothing prevents you from taking the Euclidean metric and the usual projection to the unit disk in $\mathbb R^2$ and introducing some crazy linear structure there using some wild bijection to, say, $\mathbb R^{12}$.
@fedja I guess that the ambient space $X$ has to be a normed space.
Does this answer your question? https://math.stackexchange.com/a/4935879/29780
Let $E$ be the (incomplete) subspace of sequences in $\ell^{2}(\mathbb{R})$ having at most finitely many nonzeros terms. In [1], a subset $S$ of $E$ is constructed (by a long induction argument), which has the following properties :
$S$ is closed and nonconvex,
each point in $E$ has a unique nearest point in $S$,
the projection is continuous.
This construction is mentioned in the book by D. Braess, Non linear approximation theory (p.42-43), Springer 1986.
[1] G. Johnson,
A nonconvex set which has the unique nearest point property,
J. Approx. Theory 51 (1987), no. 4, 289-332.
|
2025-03-21T14:48:29.940694
| 2020-02-27T00:26:20 |
353651
|
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|
Stack Exchange
|
Do regular functions separate points?
Suppose $K$ is a field, $V$ is a $K$-variety (finite type reduced separated $K$-scheme), and $p,q$ are $K$-points of $V$. Must there be an open subset $U$ of $V$ containing both $p$ and $q$ and a regular function $f$ defined on $U$ such that $f(p)$ and $f(q)$ are not equal? I am willing to suppose that $K$ is perfect but I doubt this is relevant.
This would give a nice proof of something I need. I have been unable to find anything useful in the literature, but I am very much not an algebraic geometer. I have asked a few algebraic geometers with no success. This question has been asked, but not answered, on stackexchange at least twice. Someone must know the answer to this.
This is clearly true if there is an open affine containing both $p$ and $q$, which always holds in many situations (e.g. if $V$ is quasi-projective). Presumably a real algebraic geometer will give a more general answer, but I thought I'd point that out in case it suffices for what you are doing.
More generally, this is fine if $X$ embeds into a (possibly non-quasiprojective) toric variety. Hironaka's example of a nonprojective smooth proper threefold (see e.g. one of the appendices in Hartshorne's textbook) at least done carefully has the feature that there are two points without common affine neighborhood. I didn't check, but it is plausible that it gives a counterexample to your desired claim as well.
@PiotrAchinger I do not think that Hironaka's example yields a counterexample since it admits a Zariski-locally projective morphism to a projective variety.
@AndyPutman Unfortunately that's not enough. I need to cover arbitrary varieties, which seems to add difficulty.
This reminds me of Weil restriction. Roughly speaking, the Weil restriction seems to be known to exist for varieties where every finite set of points is contained in an affine open subscheme. This condition of course implies yours. So maybe one can find a counterexample to your question by looking at examples where the Weil restriction does not exist?
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2025-03-21T14:48:29.940947
| 2020-02-27T02:08:34 |
353660
|
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|
Stack Exchange
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Distributive lattices and axiom of choice
What form of the axiom of choice is equivalent (in ZF) to the statement that every distributive lattice is isomorphic to a lattice of sets?
Do you mean an arbitrary isomorphism? (Just to make sure.)
Does https://mathscinet.ams.org/mathscinet-getitem?mr=873595 solve your question?
No. The Birkhoff Representation theorem is a stronger statement.
According to Wikipedia the prime ideal theorem for distributive lattices is equivalent to the Boolean prime ideal theorem. (Look in the "Further prime ideal theorems" section.)
These certainly imply the statement. But how about AC(2), the axiom of choice for 2-element sets? (Suggested to me by a well known author.)
@Fred: That makes no sense to me, honestly.
The Boolean prime ideal theorem is trivially a special case of the prime ideal theorem for distributive lattices (the reverse implication requires a bit of work). So no, the latter is not provable from just AC(2).
Also, note that by passing to quotient algebras, the Boolean PIT is equivalent to what Wikipedia calls the weak PIT: every nontrivial Boolean algebra has a prime ideal. The more general statement that every nontrivial distributive lattice has a prime ideal follows from the existence of an arbitrary embedding $f$ of the lattice in a $\mathcal P(X)$, as ${a:u\notin f(a)}$ is a prime ideal for any fixed $u\in X$. Thus, the statement that every distrib lattice is isomorphic to a lattice of sets is already equivalent to the Boolean PIT.
I mean, any fixed $u\in\bigcup_af(a)\setminus\bigcap_af(a)$.
This shows why the original answer (now checked) is correct. The axiom AC(2) is strictly weaker. See for example the book book by Herrlich "Axiom of Choice".
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2025-03-21T14:48:29.941455
| 2020-02-27T02:51:13 |
353661
|
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Stack Exchange
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Understanding the $\text{SL}_{2}$-orbit theorem
I am trying to understand a meaning of Schmid's $\text{SL}_{2}$-orbit theorem. The nilpotent orbit theorem now seems quite clear to me (both its proof and why one would want to consider such theorem), but as a first-timer I still cannot quite understand what $\text{SL}_{2}$-orbit theorem tries to say. It seems to say that you can pick a more specified nilpotent orbit satisfying all those conditions which are still mysterious to me. I tried to find some other references other than Schmid's paper (not that it is not well-written, it is just too dense for me) but I failed. I hope someone could explain an intuition behind $\text{SL}_{2}$-orbit theorem. Also, it would be great if a proof could be sketched, as I would ultimately want to understand the proof of it.
Have you seen the exposition in Griffiths-Schmid (1975, pp. 108-114)? “In very general terms [it] says the following: given any one-parameter family of Hodge structures which becomes singular, one can equivariantly embed a copy of the upper half plane in the classifying space, such that (... ...)” “Although the proof of the theorem is technical, its basic idea can be described in simple terms. We shall do so below, in the hope that this may motivate and clarify the statement”.
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2025-03-21T14:48:29.941563
| 2020-02-27T04:03:28 |
353663
|
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Stack Exchange
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Property of a semisimple Lie algebra over complex number field
Is the following property true？
Let $L$ be a finite-dimensional Lie algebra over $\mathbb{C}$. Then $L$ is semisimple if and only if for every $x \in L$, there exists $y, z\in L$, such that $x=[y, z]$.
Remark. For Lie algebra of type $A$, I know that this result is true. However, I do not know other cases.
The Lie algebra $\mathfrak{sl}_2(\mathbf{C})\ltimes\mathbf{C}^2$ is not semisimple but every element is a commutator (it has only 5 orbits under the action by automorphisms of $\mathrm{GL}_2(\mathbf{C})\ltimes\mathbf{C}^2$, so this is quite easy).
It seems anyway that you're primarily interested in the direct implication. Yes, every element is a Lie commutator in a semisimple Lie algebra. This is already asked and answered on MathSE: https://math.stackexchange.com/questions/769881/.
@YCor Thank you.
PS in my first comment I should have said "only 5 orbits modulo nonzero scalar multiplication"
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2025-03-21T14:48:29.941657
| 2020-02-27T04:40:19 |
353665
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|
Stack Exchange
|
Reference request: Projection operators in metric spaces
Given a metric space $(X,d)$ and a subset $S\subset X$, the projection $P_S$ onto $S$ is well-defined as a set valued function. I am interested in learning more about properties of these projections such as uniqueness, continuity, proximinality, etc.
Unfortunately, almost every paper I can find on this topic considers the special case where $X$ is a Banach space. I am interested specifically in the case where $X$ is a general metric space. Any pointers to papers or monographs would be appreciated.
Note. Apparently Ivan Singer has some papers from the 1960s on this, but I can’t track down the full text of the papers online. For example:
[1] Some remarks on approximative compactness. Rev. Roum. Math. Pures Appl. 9(1964), 167-177
References on metric spaces. Here are a couple papers I have found specifically regarding metric spaces:
[2] Khalil, R. (1988). Best approximation in metric spaces. Proceedings of the American mathematical society, 103(2), 579-586.
[3] Gupta, S., & Narang, T. D. (2017). STRONG PROXIMINALITY IN METRIC SPACES. Novi Sad J. Math, 47(2), 107-116.
References on Banach spaces. Just in case, here are useful some references on the Banach space version.
[4] D. D. Repovš and P.V. Semenov, Continuous Selections of Multivalued Mappings, Kluwer Academic Publishers, Dordrecht 1998.
http://www.pef.uni-lj.si/repovs/knjige/kazalo_book.htm
See Chapter C.6 on metric projections
[5] Alber, Y. I. (1996). A bound for the modulus of continuity for metric projections in a uniformly convex and uniformly smooth Banach space. journal of approximation theory, 85(3), 237-249.
[6] Deutsch, F., & Lambert, J. M. (1980). On continuity of metric projections. Journal of Approximation Theory, 29(2), 116-131.
I'm not sure if this is what you want, but closest-point projection is coarsely well-defined (ie the image sets are of bounded diameter) when the metric space is "Gromov-hyperbolic". This is used a great deal in the geometric-group-theory literature.
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2025-03-21T14:48:29.941804
| 2020-02-27T06:23:38 |
353668
|
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Stack Exchange
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Proof for an extension of Azuma's inequality
I am trying to understand a part of the proof of an extension of Azuma's inequality, where there is a small failure probability, as it appears in proposition 34 in "Random matrices: universality of local spectral statistics of non-hermitian matrices" by Terence Tao and Van Vu.
Here's the url for Arxiv:
https://arxiv.org/pdf/1206.1893.pdf
To my understanding, the basic idea for the proof is that from the original function $Y$ and a martingale sequence ($E[Y|\xi_1,\dots,\xi_{i}])_{i=1}^n$, a modified function $Y'$ and martingale sequence ($E[Y'|\xi_1,\dots,\xi_{i}])_{i=1}^n$ are constructed, by alternating $Y$ on "bad" sets where martingale difference terms are big.
I am having a hard time understanding why the new martingale sequence based on $Y'$ should have the bounded difference, i.e.
$|E[Y'|\xi_1,\dots,\xi_{i}]-E[Y'|\xi_1,\dots,\xi_{i-1}]| \leq \alpha_i$ a.s.
Any input will be very much appreciated.
Thanks!
The statement "$Y'$ satisfies the condition of Azuma’s inequality" is incorrect in general if it is supposed to mean, for instance, that $C'_i\le C\alpha_i$ for some real constant $C$ not depending on the distribution of $Y$, where $C'_i=C'_i(\xi)$ is defined similarly to $C_i(\xi)$ in definition (4.1) on page 28 in the linked paper but with $Y'$ in place of $Y$:
$$C'_i(\xi):=|E(Y'|\xi_1,\dots,\xi_i)-E(Y'|\xi_1,\dots,\xi_{i-1})|.$$
Indeed, suppose that $n=1$, $\xi_1=Y$, $P(Y=n)=\frac1{n+1}=1-P(Y=-1)$ for natural $n\ge2$, so that $EY=0$. Let $\alpha_1=2$. Then $C_1=|E(Y|\xi_1)-EY|=|Y|$, $B_1=\{C_1\ge2\}=\{Y=n\}$, $Y'=1_{B_1}\,E_{B_1}Y+1_{B_1^c}\,Y=1_{Y=n}\,n-1_{Y=-1}$, $C'_1=|Y'|=1_{Y=n}\,n+1_{Y=-1}$, which is not bounded by any constant not depending on the distribution of $Y$.
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2025-03-21T14:48:29.941937
| 2020-02-27T07:32:32 |
353672
|
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Stack Exchange
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Largest subset not spanning the span
Let $S=\{c_1,\dots,c_n\}$ be a set of vectors in $\mathbb{R}^M$. Is the below problem studied in literature?
$$\max\limits_{S'\subset S} \vert S' \vert $$
$$s.t. dim(span(S')) < dim(span(S))$$
which is to find the cardinality of the largest subset which does not span the span of S.
I won't be surprised if this is NP-hard.
I asked this question a while ago: https://mathoverflow.net/q/339819/9924 There were no replies, and somehow I convinced myself that no reasonable answer exists.
This seems to be a matroid theory question. If you let $S$ be the ground set of the matroid and you let $r$ be the rank of the matroid (i.e. $\dim \mathrm{span}(S)$), then your question amounts to finding the largest flat of rank $r-1$.
For reference, I would recommend Oxley's Matroid Theory.
That said, one generally cares more about finding the entire lattice of flats or a maximal chain of flats than a particular one. But to answer your question, flats, in general, are of particular interest in matroid theory. There are a number of tools for finding flats as well.
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2025-03-21T14:48:29.942044
| 2020-02-27T08:02:40 |
353673
|
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Stack Exchange
|
Maximal subgroups of odd index in $\mathrm{PSL}(3,q)$
Let $G = \mathrm{PSL}(3,q)$ for $q$ odd. I am trying to understand a question that involves understanding the subgroups that contain a Sylow $2$-subgroup, and in particular, are subgroups of odd index in $G$. I need to find a complete description of the maximal subgroups of odd index in the group $G = \mathrm{PSL}(3,q)$
Is this maximal proper (subgroups of odd index) or (maximal proper subgroups) of odd index?
The subgroups of ${\rm PSL}_3(q)$ for odd $q$ were first enumerated by H.H. Mitchell in 1911. (The case $q$ even was done by R.W. Hartley in 1925/6.)
Table 8.3 of the book "The Maximal Subgroups of Low-Dimensional Finite Classical Groups" by Bray, Holt and Roney-Dougal provides a convenient list. Using that it is not hard to answer your question.
For $q$ odd, the maximal subgroups of ${\rm SL}_3(q)$ of odd index are as follows:
(i) Two classes of maximal parabolic subgroups with structure ${\rm E}_q^2:{\rm GL}_2(q)$ and index $q^2+q+1$, where ${\rm E}_q$ denotes an elementary abelian group of order $q$ (the additive group of the field). These two classes are interchanged by the graph (inverse-transpose) automorphism of ${\rm SL}_3(q)$.
(ii) When $q \equiv 1 \bmod 4$, we have one class of imprimitive subgroups with structure $(q-1)^2:S_3$.
(iii) When $q = q_0^r$ for some odd prime $r$, we have $\gcd(\frac{q-1}{q_0-1},3)$ classes of subgroups with the structure ${\rm SL}_3(q_0).\gcd(\frac{q-1}{q_0-1},3)$. When there are three such classes, they are all conjugate under a diagonal outer automorphism of ${\rm SL}_3(q)$.
The maximal subgroups of odd index in finite simple groups were classified in Liebeck and Saxl - The primitive permutation groups of odd degree and independently in Kantor - Primitive permutation groups of odd degree, and an application to finite projective planes. In some cases the the subgroups listed need extra conditions to guarantee that they actually are of odd index. An explicit if and only if statement is given in Maslova - Classification of maximal subgroups of odd index in finite simple classical groups: Addendum.
A pretty complete and accessible description of this can be found in a survey article of Oliver King. In addition to the link, here's the citation info:
King, Oliver H., The subgroup structure of finite classical groups in terms of geometric configurations., Webb, Bridget S. (ed.), Surveys in combinatorics 2005. Papers from the 20th British combinatorial conference, University of Durham, Durham, UK, July 10–15, 2005. Cambridge: Cambridge University Press (ISBN 0-521-61523-2/pbk). London Mathematical Society Lecture Note Series 327, 29-56 (2005). ZBL1107.20035.
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2025-03-21T14:48:29.942232
| 2020-02-27T11:39:54 |
353681
|
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Stack Exchange
|
Examples of complicated parametric Jordan curves
For test purposes I need parametric Jordan curves that are complicated in the sense of having many inflection points and ideally no symmetries.
When doing online search I always land at complex analysis and curves related to conformal maps.
Explicit formulas or recipes for generating such curves would be great; even better if sample images could be provided.
I need those kind of test curves for checking an algorithm for generating characteristic functions from closed curves and for checking the approximability of those functions.
You can take any continuous (or as smooth as you desire) function $f:[0,2\pi]\to(0,\infty)$ with equal values at the end points and plot it in polar coordinates.
@Wojowu those functions would be starshaped and it should be noted that $f$ shouldn't have sign changes in $(0,\pi)$
They would indeed be starshaped, but not knowing exactly what it is that you want to do with them, I don't know how good of a suggestion that was (hence I posted it as a comment and not as an answer). I think it should be noted that a function taking values in $(0,\infty)$ cannot have any sign changes :)
I don't see where you need any additional restrictions. For $f$ in values in $(0,\infty)$ as I wrote you have no sign changes and no self intersections.
@Wojowu agreed; I didn't look careful enough. Regarding what I need the curves for: I have an idea for generating a smooth function for e.g. visualising all Euclidean distances between pairs of curve points and I want to put that idea to test with non-trivial examples; that idea would also work in higher dimensions.
Some examples and references are mentioned here Examples of plane algebraic curves. You can find many Jordan curves in the family $e^{it}+re^{int}, 0\leq t\leq 2\pi,$ by choosing parameters properly.
To generalize this, take any polynomial $P$ which is univalent (=injective) in the closed unit disk. The image of the unit circle is a Jordan curve. By Riemann's theorem these curves are dense in the set of all Jordan curves, so it
can be arranged that they have plenty of inflection points. A huge supply of univalent polynomials is given by the formula $P(z)=z+\epsilon Q(z)$, where $Q$ is any polynomial and $\epsilon<1/\max_{|z|=1}|Q'(z)|$. These curves depend on $\epsilon$ as parameter, and tend to the unit circle when $\epsilon\to 0$.
Another large class of examples are certain lemniscates, they are not given as parametrized
curves but their parametrization sometimes can be obtained. Example: $\{ z:|z^n+1|=k\}$, where
$k>1$.
In general, a lemniscate is a level set of a complex polynomial $P(z)$.
It is Jordan, when the level $k$ is larger than all critical values of the polynomial. (For other $k$ not equal to moduli of critical values, they
are disconnected unions of Jordan curves).
They may have plenty of inflection points. To obtain a parametrization, you have to be able to invert the polynomial, if $f$ is the inverse, then the parametrization
is $f(ke^{it}), 0\leq t\leq 2\pi d$, $d=\deg P$. Like the previous examples they come
in continuous families; when $k$ is large, they resemble circles.
A theorem of Hilbert says that every Jordan curve can be approximated
by lemniscates so the variety of their shapes is unlimited, but of course most
of them cannot be explicitly parametrized. Still those which can be give plenty of examples.
Remark. It is not a surprise that closed curves in the plane are mostly mentioned
in connection with functions complex variables. Complex variables give the most
convenient way to describe and study them.
My answer is similar to the one given by Prof. Eremenko, but with a less classical flavor. The "supercurves", a family of parametric plane curves described by Superformulas and introduced by Johan Gielis around the year 2000, describe many complex plane shapes, including "natural" ones like leaves, flakes and the like: the three dimensional generalization of these object describe "natural" 3D shapes as well.
[1] J. Gielis, D. Caratelli, Y. Fougerolle, P. E. Ricci, I. Tavkelidze, T. Gerats, "Universal Natural Shapes: From Unifying Shape Description to Simple Methods for Shape Analysis and Boundary Value Problems", PLoS ONE, Public Library of Science, (2012), 7, e29324.
An airplane fellow-passenger in the early 1980s(?) (his last name was probably Brown -- sorry, I can't be $100\%$ sure) shared with me his own idea; I believe that he was an engineer or designer, certainly not a professional mathematician.
For every $(a\ b)\in\mathbb Z^2\ $ draw two quarter-circles of radius $\ \frac 12,\ $ inside the square $\ [a;a+1]\times[b;b+1]\ $ such the centers are either
$$ (a\ b)\qquad\qquad\qquad(a\!+\!1\ \ b\!+\!1) $$
or
$$ (a\!+\!1\ \ b)\qquad\qquad\qquad(a\!+\!1\ \ b) $$
Each (countable) choice will provide a 1-dimensional submanifold of $\ \mathbb R^2.\ $ For most of the random choices there will be complicated curves.
Remark 1: It's easy to replace the said curves by similar but smooth (infinitely differentiable) curves.
Remark 2: Following my fellow passenger idea, I defined a similar
hexagonal version, where instead of two quarter-circles
(selected in the two ways inside the integer squares),
one would draw three $\ \frac{2\cdot\pi}3-$arcs or
two such arc and an interval which connects the
middle-points of the opposite edges of the hexagons
of the hexagonal lattice.
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2025-03-21T14:48:29.942578
| 2020-02-27T12:02:32 |
353682
|
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Stack Exchange
|
What is the winning strategy in this pebble game?
Consider the following two-player pebble game. We have finitely
many stones on a finite linear track of squares. We take turns, and
the allowed moves are:
move any one stone one square to the left, if that square is empty, or
remove any one stone, or
remove any two adjacent stones.
Whoever takes the last stone wins.
Question. What is the winning strategy? And which are the
winning positions?
The game will clearly end always in finitely many moves, and so by
the fundamental theorem of finite games, one of the players will
have a winning strategy. So of course, I know that there is a
computable winning strategy by computing with the game tree, and we have a computable algorithm to answer any instance of the question. What I am hoping for is that there will be a simple-to-describe winning strategy.
This is what I know so far:
Theorem. It is a winning move to give your opponent a position
with an even number of stones, such that the stones in each successive pair stand
at even distance apart.
By even-distance, I mean that there are an odd number of empty
squares between, so adjacent stones count as distance one, hence
odd. Also, I am only concerned with the even distance requirement within each successive pairs, not between the pairs. For example, it
is winning to give your opponent a position with stones at
..O...OO.O....O.....O...........
We have distance 4 in the left-most pair, distance 2 in the next pair, distance 6 in the third pair, ignoring the distances in front and between the pairs.
Proof. I claim that if you give your opponent a position like
that, then he or she cannot give you back a position like that, and
furthermore, you can give a position like that back again. If your
opponent removes a stone, then you can remove the other one in that
pair. If your opponent moves the lead stone on a pair, then you can
move the trailing stone. If your opponent moves the trailing stone
on a pair, then either you can move it again, unless that pair is
now adjacent, in which case you can remove both. And if your
opponent removes two adjacent stones, then they must have been from
different pairs (since adjacent is not even distance), which would
cause the new spacing to be the former odd number plus another odd
number plus 2, so an even number of empty squares between, and so
you can move the trailing end stone up one square to make an odd
number of empty squares between and hence an even distance between
the new endpoints. Thus, you can maintain this even-distance
property, and your opponent cannot attain it; since the winning
move is moving to the empty position, which has all even distances,
you will win. $\Box$
What I wonder is whether there is a similarly easy to describe
strategy that solves the general game.
May I ask about the origins of this game, is it related to other mathematics or did it arise as an isolated question?
I was analyzing another game, a variant of my game Buckets of Fish, and had reduced it to this pebble game. The variant game was just like Buckets of Fish, except that you can only add fish to the immediately adjacent bucket. If you think only of the odd-bucket positions, you have the pebble game. http://jdh.hamkins.org/buckets-of-fish/
The positions which are a win for the second player are those with:
an even number of pebbles in odd-numbered squares, and
an even number of pebbles in even-numbered squares.
Indeed, from a position in this set $P$, any move will be to a position not in that set, whereas from a position not in that set one can always make a move to a position in that set (if only the number of pebbles in the odd-numbered squares is odd, you can remove one, similarly for the even-numbered, and if both are odd, choose any pebble — that is not at the very leftmost square — and either more it to the left if that square is unoccupied or remove it along with the one to the left if it is).
So, systematically moving to a position in $P$ (as described in the parenthesis in the previous paragraph) provides a winning strategy provided one starts with a position not in $P$.
Fantastic! This seems to work beautifully. I'll mull it over, and probably accept later.
I am kicking myself for missing this solution, since I had looked at many different variations on parity conditions, none of which worked. Your criteria both generalizes and simplifies the partial solution I describe in the question. Excellent.
We can go further and find the nim-value of any position of this game.
Theorem. The nim-value of a position in this game is the Nim-sum of the contributions of its pebbles, where
Pebbles in odd-numbered squares contribute $1$.
Pebbles in even-numbered squares contribute $2$.
(Here we count the leftmost square as the $1^{st}$ square.)
Corollary. As in Gro-Tsen's answer, the winning positions (with nim-value $0$) are the ones with both an even number of pebbles in odd-numbered squares and an even number of pebbles in even-numbered squares.
Corollary. All positions have nim-value $0, 1, 2,$ or $3$.
The proof relies on the following idea, which is encoded in the lemma below.
Key Insight. A position in this pebble game is equivalent to the sum of an individual game for each pebble with its own private track.
Lemma. Given a position $P$ of pebbles $p_1, p_2, ... p_n$, consider the games $P_i$ which each consist of only the single pebble $p_i$ in the same square. Then $P$ is equivalent to $\sum_i P_i$.
Proof. To check this, we need only show that the game $P + \sum_i P_i$ is a zero game, by demonstrating a winning strategy for the second player. So consider the possible moves for the first player in the game $P + \sum_i P_i$:
If the first player removes two adjacent pebbles $p_i$ and $p_{i+1}$ in $P$, the second player can respond by moving the pebble in $P_{i+1}$ to the left, leaving behind two copies of $P_i$, which nim-cancel.
If the first player removes any one pebble, the second player responds by removing the corresponding pebble.
If the first player moves a pebble to the left in $P$, the second player responds by moving the pebble to the left in $P_i$.
If the first player moves a pebble to the left in a $P_i$, the second player responds by moving the corresponding pebble to the left in $P$. If that move is blocked by a pebble directly to the left, that's fine -- the second player instead removes those two pebbles, noting that there are now two copies of $P_{i-1}$ that nim-cancel out.
Proof of Theorem. Because of the Lemma, we need only consider the case of a single pebble in position $i$ (where position $1$ is the leftmost square). By induction on $i$,
There is always the option to remove the pebble, an option with nim-value $0$.
If $i$ is odd, there may also be the option to move the pebble to the left, which is an even-numbered square with nim-value $2$. Thus the nim-value in the odd case is $mex(0,2) = 1$.
If $i$ is even, there is always also the option to move the pebble to the left, which is an odd-numbered square with nim-value $1$. Thus the nim-value in the even case is $mex(0,1) = 2$.
How to Evaluate Positions.
Example 1. The nim-value of the position in the question
. . O . . . O O . O . . . O O . . . . . O
is found quickly by seeing that a pebble in an odd-numbered position contributes nim-value 1, and a pebble in an even-numbered position contributes nim-value 2. So using this mask:
. . O . . . O O . O . . . O O . . . . . O
1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1
yields the Nim-sum
1 +1+2 +2 +2+1 +1
which equals 2, so removing any of the even-numbered pebbles (the "2s") is a win.
Example 2. You are playing the sum of the following two positions:
O . O . . . O O . O + O . O O
Evaluate each one with the mask:
O . O . . . O O . O + O . O O
1 2 1 2 1 2 1 2 1 2 1 2 1 2
1 +1 +1+2 +2 + 1 +1+2
This has a Nim-sum of 3, so moving any pebble to the left or removing any two adjacent pebbles (in either component!) is a win.
Very good; this is very helpful.
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